design of transmission line between alkudmi and medical city of jazan
TRANSCRIPT
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Jazan university
College of Engineering
Electrical Engineering Department
Done By :
M.A.ALAJAM 201013054
M.D.ALAKHRASH 201010115
J.M.ALLOLI 201013027
O.A.SAADI 201010823
A.M.QASEM 201010293
Supervisors:
Prof. Dr .Atef Elemary
Dr. Ehab Salim
This project was carried out in the year 2014
Design of Transmission Line between ALKUDMI
and Medical City of Jazan
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[Introduction]Chapter 1
2
[بطون
ن
كم ج خر
ه
كمو
ل ج و
ئي ش
ون
شكرون أه تكم هكم
ة
فئد
ا
ر و
بص
ا
ع وه
(اح87)[اس
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[Introduction]Chapter 1
3
Acknowledgment
In the first we thanks Allah. Our sincere thanks go to Prof Dr. Atef
Elemary, for his guidance, suggestions, continuous encouragement during
the progress of this project . Also, we are thanks Dr. Ehab Salim for his,
advice, kind assistance and enormous patience throughout this work.Moreover, my thanks go to the staff of the electrical department for their
support over the past years.
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[Introduction]Chapter 1
4
Page:Contents :
7Chapter1: Introduction81.1 Introduction
91.2 equivalent electrical network of Jazan10Chapter2:Electrical Performance of Transmission Line112.1 Introduction
112.2 Type of conductor
122.3 Transmission Line Parameters
12 2.3.1 Resistance
13 2.3.2 Inductance and Inductive Reactance
142.3.3 Transmission Line Configuration
14 2.3.4 Inductance of Three Phase T.L with Symmetrical Spacing
14 2.3.5 Inductance of Three Phase T.L with Asymmetrical Spacing
15 2.3.6 Transpose Transmission Line 16 2.3.7 Inductance Of Three Phase Double Circuit Lines
17 2.3.8 Bundled Conductors
17 2.3.9 Capacitance of Transmission Line
172.3.10 Capacitance of Three Phase Transmission Lines
18 2.3.11 Capacitance of Three Phase Double Circuit
19 2.3.12 Capacitance Due to Earth’s Surface
212.4 Classification of Overhead Transmission Lines
21 2.4.1 Important Terms
22 2.4.2 Characteristics and performance of transmission lines
22 2.4.3 Short Transmission Lines
23 2.4.4 Medium Transmission Lines 23 2.4.5 Long Transmission Lines
242.5 Calculations of our Transmission Line
24 2.5.1 Calculation Of Inductance
25 2.5.2 Calculation Of Capacitance
26 2.5.3 Performance calculation
262.5.4 Efficiency and Regulation
272.5.5 Load Curve
27 2.5.6 Load Ability Curve
282.6 Phenomenon of Corona
28 2.6.1introduction28 2.6.2 Corona affected by the following factors
28 2.6.3 Advantage of Corona
28 2.6.4 Disadvantage of Corona
29 2.6.5 Visual critical voltage
30Chapter3 :Mechanical Design of Transmission Line
313.1 Introduction
313.2 Factors affecting mechanical design of overhead lines
323.3 Nature of line route
323.4 Mechanical loading:
32 3.4.1 Stresses
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[Introduction]Chapter 1
5
33 3.4.2 Elasticity
34 3.4.3 Wind pressure
353.5 Clearance:
35 3.5.1 Horizontal separation of conductors from each other
363.6 Type of supporting structure
373.7 Insulator:
383.8 Sag and Tension analysis
39 3.8.1 Effect of change in temperature
40 3.8.2 Calculation of line sag and tension
41 3.8.3Supports at same level
42 3.8.4 Effect of wind
43 3.8.5 Calculation of line sag and tension of our project : 4Chapter4: Power Flow Analysis
464.1 Introduction
464.2 Necessity for Power Flow Studies
464.3 Bus Classification
46 4.3.1 Slack or Swing Bus
46 4.3.2 Generator or Voltage Controlled Bus
47 4.3.3 Load Bus
474.4 Load Flow Equations:
474.5 Techniques of Solution
48 4.5.1 Gauss - Seidel Iterative Method of Load Flow Solution
48 4.5.2 The Newton-Raphson Power-Flow Solution
504.6 Load flow by using Matlab program
514.7 Load flow by using Power world Simulator Program
52Chapter 5: Fault System Analysis
535.1 Introduction455.2 Objective
535.3 Types of common faults
535.4 Symmetrical Short circuit (3-phace)
545.5 Unsymmetrical Short Circuit
54 5.5.1 Symmetrical Components
565.5.2 Single Line to ground (L-G)
57 5.5.3 Line-to-line fault
585.5.4
Double Line-to-Ground Fault
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[Introduction]Chapter 1
6
595.6 Short circuit calculations
61Chapter6: Power System Stability
626.1 Introduction :
626.2 Power System Stabilizers
63 6.2.1 PSS gain, time constant and block diagram
6.2.2 Block diagram of PSS in Matlab Simulink Program
65Chapter7: Distribution System for
667.1 Introduction
667.2 Causes of voltage drops
66 7.2.1 There are four fundamental causes of voltage drop
677.3 Application Of Capacitors to Distribution Systems :
67 7.3.1 Basic Definitions
677.4 Power Factor Improvement:
67 7.4.1 Fixed Versus Switched Capacitors
68 7.4.1.1 Sizing and Location of Capacitors
69 7.4.1.2 Effect of Shunt Capacitors on Radial Feeders
70 7.4.2 Fixed and Switched Capacitor Applications
707.5 System Benefits :
07 7.5.1 Effects Of Series And Shunt Capacitors
71 7.5.2 series Capacitors
72 7.5.3 Shunt Capacitors
73 7.5.4 A Mathematical Procedure to Determine the Optimum Capacitor
Allocation
747.6 loss reduction due to capacitor allocation
00 7.6.1 Calculation Reduction of Power losses after add capacitor for Residential
area
18 7.6.1 Calculate the voltage drop per phase
83Chapter8: Conclusions
14Appendix
88Reference
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[Introduction]Chapter 1
7
Chapter 1
Introduction
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[Introduction]Chapter 1
8
1.1 Introduction :
The main function of our project is to study of design transmission lines between Alkudmi and
Medical city of Jazan , this study pass through a some stages electrical and mechanical performances
of this transmission lines , and analysis the electrical power flow in the network of Jazan city ,short
circuit calculations of Jazan network that found in the project. Then , we estimate the stability of
our transmission lines . Finally we study distribution system of Residential district.
Where the electrical performance contain common types of conductors and transmission line
parameters .Also, we study the efficiency of electrical power between sending and receiving points.
The mechanical performance in this project is to study construction and how supports towers to
withstand at all expected load and also against the nature powers. Also, the mechanical designs the
materiel chosen for conductor to withstand against uncertain whether conditions and forces that will
subjected to it.Also, calculation sag and tension of the transmission lines.
The power flow studies are necessary for planning, economic operation scheduling and exchange of
power between utilities. Power flow studies used to determine the bus voltage magnitude and angle,
power flow through line, losses and generation capacity. Buses are classifying into three category
slack bus, load bus and voltage controlled bus.
Fault studies form an important part of power system analysis and the problem consists of
determining bus voltage and line current during faults. Fault studies are used for proper choice of
circuit breakers and protective relaying.
Power system stability is the ability of the system to return to normal operation after having beensubjected to disturbance in the system. Instability means a condition denoting to loss of synchronism
of synchronous machine or falling out of step. Therefore the state of equilibrium or stability of a
power system commonly alludes to maintaining of synchronous operation of the system. Three types
of stabilities are of concern steady state, dynamic and transient stability.
The study of distribution system of Residential district is to determine the voltage drop by
conductors or connections of leading to the electrical load .Voltage drop is caused by material , wire
size and current being carried .Also, we study applications of the capacitance and how to improve
power factor. And how to deal with mathematical formulas of it.
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[Introduction]Chapter 1
9
The Figure (1.1) shows the equivalent electrical network of Jazan .
Figure (1.1) : equivalent electrical network of Jazan
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Electrical Performance of Transmission LineChapter 2
[01]
Chapter 2
Electrical Performance of
Transmission Lines
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Electrical Performance of Transmission LineChapter 2
[00]
Electrical Performance of Transmission Lines
2.1 Introduction :
There are two types of transmission line conductors: overhead and underground. Overhead
conductors, made of naked metal and suspended on insulators, are preferred over underground
conductors because of the lower cost and easy maintenance. Also, overhead transmission lines use
aluminum conductors, because of the lower cost and lighter weight compared to copper conductors,
although more cross-section area is needed to conduct the same amount of current.
2.2 Type of conductor:
There are different types of commercially available aluminum conductors:
1-
All Aluminum Conductor AAC2-
All Aluminum Alloy Conductor AAAC
3- Aluminum Conductor Alloy Reinforced ACAR
4- Aluminum Conductor Steel Reinforced ACSR
Figure 2.1 shows an examples of different types of conductors .
Figure (2.1): Different types of conductors
•
ACSR conductor is most commonly used. Due to :
1. It cheaper than copper conductors of equal resistance .
2.
Corona losses are reduced due to large diameter of the conductor.
3. It has superior mechanical strength & hence span of large lengths which results in small
number of supports
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Electrical Performance of Transmission LineChapter 2
[0]
Figure 2.2 shows an example of aluminum and steel strands combination.
Figure. (2.2) : Stranded aluminum conductor with stranded steel core (ACSR). (7 steel strands and 30 aluminum strands).
2.3 Transmission Line Parameters
An overhead transmission line consists of a group of conductors running parallel to each other and
carried on supports which provide insulation between the different conductors and between each
conductor and earth. A transmission line has four parameters. Resistance, inductance, capacitance
and shunt conductance. The shunt conductance accounts for leakage currents flowing across
insulators and ionized pathways in the air. The leakage currents are negligible as compared to the
current flowing in the transmission lines. The series resistance causes a real power loss in the
conductor. The resistance of the conductor is very important in transmission efficiency evaluation
and economic studies. The power transmission capacity of the transmission line is mainly governed by the series inductance. The shunt capacitance causes a charging current to flow in the line and
assumes importance for medium and long transmission lines. These parameters are uniformly
distributed throughout but can be lumped for the purpose of analysis on approximate basis.
2.3.1 ResistanceThe AC resistance of a conductor in a transmission line is based on the calculation of its DC
resistance. If DC current is flowing along a round cylindrical conductor, the current is uniformly
distributed over its cross-section area and its DC resistance is evaluated by equation (2.1) :
(2.1)
where conductor resistivity at a given temperature (Ω-m)
conductor length (m)
A=conductor cross-section area (m2)
If AC current is flowing, rather than DC current, the conductor effective resistance is higher due to
frequency or skin effect.
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Electrical Performance of Transmission LineChapter 2
[0]
A- Frequency EffectThe frequency of the AC voltage produces a second effect on the conductor resistance due to the non-
uniform distribution of the current. This phenomenon is known as skin effect. As frequency
increases, the current tends to go toward the surface of the conductor and the current density
decreases at the center. Skin effect reduces the effective cross-section area used by the current, and
thus, the effective resistance increases. Also, although in small amount, a further resistance increaseoccurs when other current-carrying conductors are present in the immediate vicinity. A skin
correction factor k, obtained by differential equations and Bessel functions, is considered to
reevaluate the AC resistance. For 60 Hz, k is estimated around 1.02
R AC=R AC k (2.2)
Other variations in resistance are caused by
• . Temperature
• . Spiraling of stranded conductors
• . Bundle conductors arrangement
B- Temperature Effect
The resistivity of any conductive material varies linearly over an operating temperature, and
therefore, the resistance of any conductor suffers the same variations. As temperature rises, the
conductor resistance increases linearly, over normal operating temperatures, according to the
following equation:
(2.3)
where R 2 = resistance at second temperature t2
R 1 = resistance at initial temperature t1T = temperature coefficient for the particular material (Co)
Resistivity () and temperature coefficient (T) constants depend upon the particular conductormaterial.
2.3.2 Inductance and Inductive ReactanceA current-carrying conductor produces concentric magnetic flux lines around the conductor. If the
current varies with the time, the magnetic flux changes and a voltage is induced. Therefore, an
inductance is present , defined as the ratio of the magnetic flux linkage and the current. The magnetic
flux produced by the current in transmission line conductors produces a total inductance whose
magnitude depends on the line configuration.
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Electrical Performance of Transmission LineChapter 2
[0]
2.3.3 Transmission Line Configuration
Figure(2.3) shows the transmission line configuration for 132 k v
2.3.4 Inductance of Three Phase T.L with Symmetrical Spacing
Figure (2.4 )shows the conductors of a three phase transmission line with symmetrical spacing.Radius of conductor in each phase is r .
(2.4)
Where :
(2.5)
2.3.5 Inductance of Three Phase T.L with Asymmetrical Spacing
In actual practice, the conductors of a three phase transmission line are not at the corners of an
equilateral
triangle because of construction considerations. Therefore
with asymmetrical spacing,even with balanced currents , the flux linkages and inductance of each phase are not the same. A
different inductance in phase, resulting in unbalanced receiving-end voltages even when sending-
Figure (2.4) :Three Phase line with
symmetrical spacing
Figure (2.3): Conductor arrangements for 380 kV
overhead double circuit
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Electrical Performance of Transmission LineChapter 2
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end voltages and line currents are balanced. Figure 2.5 shows the conductors of a three phase
transmission line .
The above equations (2.6) , show that the phase inductances are not equal , due to mutual inductance
they contain imaginary terms .
2.3.6 Transpose Transmission Line
As mentioned in the previous equations, asymmetrical spacing gives complex values of phase
inductances, which makes the study of power system difficult. However, one way to regain symmetry
in good measure and obtain a per phase model by exchanging the positions of the conductors at
regular intervals along the line such that each conductor occupies the original position of every other
conductor. Such an exchange of conductor positions is called transposition. The transposition isusually carried out at switching stations. A complete transposition cycle is shown in Fig. 2.6. This
arrangement causes each conductor to have the same average inductance .
Figure. (2.5): Three phase line with with asymmetrical spacing. asymmetrical spacing.
(2.6)
Fig. (2.6): Transposition cycle of three-phase line.
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Electrical Performance of Transmission LineChapter 2
[0]
(2.7)
(2.8)
GS =r ' (2.9)
2.3.7 Inductance Of Three Phase Double Circuit LinesA three phase double circuit line consists of two parallel conductors for each phase. It is common
practice to build double-circuit three phase lines for greater reliability and higher transmission
capacity. To enhance the maximum transmission capability, it is desirable to have a configuration
which results in minimum inductance per phase. This is possible if mutual GMD (D m) is low and self
GMD (DS) is high.
Figure 2.7 shows the three sections of the transposition cycle of a double circuit three phase line. This
configuration gives high value of DS (Reader may try other configurations to verify that these will
lead to low DS )
To calculate the inductance, it is necessary to determine Geometric Mean Distance (GMD) and s
Geometric Mean Radius GMR .
Figure (2.7): Arrangement of conductors in three phase double circuit
(2.10)
Where :
(2.11)
(2.12)
GMR C: get from stander table of conductor code word .
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Electrical Performance of Transmission LineChapter 2
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2.3.8 Bundled ConductorsIt is economical to transmit large amount of power over long distances by EHV lines and EHV lines
are usually constructed with bundled conductors. Bundled conductors increase the self GMD and line
inductance is reduced considerably which increase the power capability of the transmission line.
Bundled conductors also reduce the corona loss, surge impedance and radio interference. The bundle
usually comprises two, three or four conductors as shown in Fig. 2.8.
o For a two conductor arrangement
√ (2.13)
o For a three conductor arrangement
√ (2.14)
o For a four conductor arrangement
√ (2.15)
2.3. 9 Capacitance of Transmission Line :Transmission Line conductors exhibit capacitance with respect to each other due to the potential
difference between them. this capacitance together with conductance forms the shunt admittance of
T.L. the conductance is the result of leakage over the surface of insulators and is negligible . When
AC voltage is applied to the T.L, the line capacitance is proportional to the length of the T.L and may
neglect for the line less than 100 Km of length.
2.3.10 Capacitance of Three Phase Transmission Lines :
Figure (2.9) shows a three phase line, each with radius r and lines are transposed.
For equilateral spacing, D12= D23 =D32 =D and Deq = D Therefore
Figure (2.8): Configuration of bundled conductors
Figure. (2.9): Three phase transmission line (fully transpose).
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Electrical Performance of Transmission LineChapter 2
[0]
(2.16)
For bundle conductors , capacitance given by
( ) (2.17)
Where DS is found in equations (2.13) ,(214) and (2.15)
3.3.11 Capacitance of Three Phase Double Circuit:
Figure (2.10): Three section of three phase double circuit
Each phase conductor is transposed within its groups. The effect of ground and shield wires are
considered to negligible. In this case per phase equivalent capacitance to neutral is :
(
) (2.18)
Where :
(2.19)
Note that:
Deq & DS will remain same for section II and section III of transposition cycle .
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Electrical Performance of Transmission LineChapter 2
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2.3.12 Capacitance Due to Earth’s Surface :
Considering a single-overhead conductor with a return path through the earth, separated a distance H
from earth’s surface, the charge of the earth would be equal in magnitude to that on the conductor but
of opposite sign. If the earth is assumed as a perfectly conductive horizontal plane with infinite
length, then the electric field lines will go from the conductor to the earth, perpendicular to the
earth’s surface ( Figure 2.11)
To calculate the capacitance, the negative charge of the earth can be replaced by an equivalent charge
of an image conductor with the same radius as the overhead conductor, lying just below the overhead
conductor (Fig. 2.12).
Figure (2.11) Distribution of electric field lines from an overhead conductor to earth’s surface.
Figure (2.12) Equivalent image conductor representing the charge of the earth.
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Electrical Performance of Transmission LineChapter 2
[1]
The same principle can be extended to calculate the capacitance per phase of a three-phase system.
Figure 2.13 shows an equilateral arrangement of identical single conductors for phases A, B, and C
carrying the charges qA, qB, and qC and their respective image conductors A', B' , and C '.
DA, DB and DC are perpendicular distances from phases A,B, and C to earth’s surface. DAA', DBB', and
DCC' are the perpendicular distances from phases A,B and C to the image conductors A',B' and C'.
As overhead conductors are identical, then r = r A= r B = r C . Also, as the conductors have equilateral
arrangement, D=DAB=DBC=DCA .
Therefore, the phase capacitance CAN, per unit length, is
That term represents the effect of the earth on phase capacitance, increasing its total value. However,
the capacitance increment is really small, and is usually neglected, because distances from overhead
conductors to ground are always greater than distances among conductors .
Figure (2.13) Arrangement of image conductors in a three-phase transmission line.
(2.20)
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Electrical Performance of Transmission LineChapter 2
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2.4 Classification of Overhead Transmission Lines
A transmission line has three constants R, L and C distributed uniformly along the whole length of
the line. The resistance and inductance form the series impedance. The capacitance existing between
conductors for 1-phase line or from a conductor to neutral for a 3-phase line forms a shunt path
throughout the length of the line. Therefore, capacitance effects introduce complications in
transmission line calculations. Depending upon the manner in which capacitance is taken into
account, the overhead transmission lines are classified as :
(i ) Short transmission lines. When the length of an overhead transmission line is upto about
50 km and the line voltage is comparatively low, it is usually considered as a short
transmission line. Due to smaller length and lower voltage, the capacitance effects are
small and hence can be neglected. Therefore, while studying the performance of a short
transmission line, only resistance and inductance of the line are taken into account.
(ii)
Medium transmission lines. When the length of an overhead transmission line is about 50-150 km and the line voltage is moderately high, it is considered as a medium transmission
line. Due to sufficient length and voltage of the line, the capacitance effects are taken into
account. For purposes of calculations, the distributed capacitance of the line is divided and
lumped in the form of condensers shunted across the line at one or more points.
(ii i) Long transmission lines. When the length of an overhead transmission line is more than
150 km and line voltage is very high, it is considered as a long transmission line. For the
treatment of such a line, the line constants are considered uniformly distributed over the
whole length of the line and rigorous methods are employed for solution.
It may be emphasized here that exact solution of any transmission line must consider the fact that the
constants of the line are not lumped but are distributed uniformly throughout the length of the line.
However, reasonable accuracy can be obtained by considering these constants as lumped for short
and medium transmission lines.
2.4.1 Important TermsWhile studying the performance of a transmission line, it is desirable to determine its voltage
regulation and transmission efficiency. We shall explain these two terms in turn.
(i )
Voltage regulation(ε). When a transmission line is carrying current, there is a voltagedrop in the line due to resistance and inductance of the line. The result is that receiving
end voltage (VR) of the line is generally less than the sending end voltage (VS ). This
voltage drop (VS −VR) in the line is expressed as a percentage of receiving end voltage VR
and is called voltage regulation.
Mathematically,
|||||| (2.21)
Obviously, it is desirable that the voltage regulation of a transmission line should be low i.e., the
increase in load current should make very little difference in the receiving end voltage.
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Electrical Performance of Transmission LineChapter 2
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(ii) Transmission efficiency(η). The power obtained at the receiving end of a transmission
line is generally less than the sending end power due to losses in the line resistance.
100
P
P)(Efficiency
S
R
(2.22)
2.4.2 Characteristics and performance of transmission lines The characteristics and performance of transmission lines. It is convenient to represent a transmission
line by the two-port network, wherein the sending-end voltage VS
and current IS
are related to the
receiving-end voltage VR
and current IR
through A, B, C and D parameters as :
[ ] [
] [ ] (2.23)
A, B, C and D are the parameters that depend on the transmission-line constants R, L, C and G. The
ABCD parameters are, in general, complex numbers. A and D are dimensionless. B has units of ohms
and C has units of Siemens . Also the following identify holds for ABCD constants.
AD-BC = 1 (2.24)
To avoid confusion between total series impedance and series impedance per unit length,
the
following notation is used :
series impedance per unit length
, shunt admittance per unit length
total series impedance
total shunt admittance
l = line length, m.
Note that the shunt conductance G is usually neglected for overhead transmission system.
2.4.3 Short Transmission Lines
As stated earlier, the effects of line capacitance are neglected for a short transmission line. Therefore,
while studying the performance of such a line, only resistance and inductance of the line are taken
into account. The equivalent circuit of a single phase short transmission line is shown in Fig. 2.14 .
Therefore the constants ABC and D are
VS VR
R XL
Figure (2.14):Single phase Short line Model
(2.25)
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Electrical Performance of Transmission LineChapter 2
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2.4.4 Medium Transmission Lines
In short transmission line calculations, the effects of the line capacitance are neglected because such
lines have smaller lengths and transmit power at relatively low voltages (< 20 kV). However, as the
length and voltage of the line increase, the capacitance gradually becomes of greater importance.
Since medium transmission lines have sufficient length (50-150 km) and usually operate at voltagesgreater than 20 kV, the effects of capacitance cannot be neglected. Therefore, in order to obtain
reasonable accuracy in medium transmission line calculations, the line capacitance must be taken into
consideration.
Therefore the constants A B C and D are :
2.4.5 Long Transmission Lines
It is well known that line constants of the transmission line are uniformly distributed over the entirelength of the line. However, reasonable accuracy can be obtained in line calculations for short and
medium lines by considering these constants as lumped. If such an assumption of lumped constants is
applied to long transmission lines (having length excess of about 150 km), it is found that serious
errors are introduced in the performance calculations. Therefore, in order to obtain fair degree of
accuracy in the performance calculations of long lines, the line constants are considered as uniformly
distributed throughout the length of the line. Rigorous mathematical treatment is required for the
solution of such lines.
Therefore the constants ABC and D are :
Figure (2.15):Single phase medium line Model
Figure (2.16):Single phase Long line Model
(2.27)
(2.26)
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Electrical Performance of Transmission LineChapter 2
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2.5 Calculations of our transmission line :
In our project we study the electrical performance of transmission line between Alkudmi and Medical
city of Jazan , we found the distance between them around 80 km , Also after communication withSaudi Company of Electric we found , their use Grosbeak type . Table 2.2 shows data of Grosbeak
(ACSR) conductor .
Table 2.2 data of Grosbeak conductor
Code
word
Cross Section Area Strading
Al/Steel
Diameter layers Approx. current-
carrying capacityTotal
(mm2)
Aluminum
Kcmil mm2 Conductor
(mm)
Core
(mm)
Grosbeak 375 363 322 26/7 25.15 9.27 2 780
Code
word
Reactance (mΩ/km) GMR
(mm)
layers 60 Hz reactance
Dm =1m
DC
25 c
AC (60 Hz) X 1 Xo
25 c 50 c 75 c
Grosbeak 91.7 92.2 101.2 110.3 194.5 0.214 7.1 0.346 0.209
Diameter = 25.15 mm =0.02515 m , radius = 12.75×10-3m length of T.L =80 km
Receiving end current (IR )= 780 A , Resistance (R) = 92.2 mΩ/km
Receiving end voltage VR =132 kv , GMR c=10.21 mm= 0.01021 m
Referred to transmission line configuration (figure 2.3) and a three sections of the transposition cycle
of a double circuit three phase line (figure 2.7) we get .
2.5.1 Calculation Of Inductance :
Figure (2.17): Conductors spacing arrangements for 132 kVoverhead double circuit
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Electrical Performance of Transmission LineChapter 2
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Da1b1= 4 m Da1c1=8 m Db1c1= 4 m
Da1b2=7.46 m Da1c2=6 m Db1c2= 7.46 m
Da1a2= 9.88 m Db1b2=6.6 m Dc1c2= 9.88 m
Therefore ;
√
( )
2.5.2
Calculation Of Capacitance :
( )
Where :
, ,
,
,
√ , √
√ √
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Electrical Performance of Transmission LineChapter 2
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2.5.3 Performance calculation :
Figure (2.18): Represent to medium T.L
Parameter of Medium T.L :
Where :
R=7.376 /km , L=0.59817 mH/km , C=0.019397
Assume PF=0.9 lag
So ,
,
√ |
| 2.5.4 Efficiency and Regulation
|
√ |
|
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Electrical Performance of Transmission LineChapter 2
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Active power at maximum load
√
√
Efficiency =
Regulation = |||||| %
2.5.5 Load Curve
Figure: (2.19) Load curve
2.5.6 Load Ability Curve
Figure: (2.20) Load Ability Curve
0 200 400 600 800 1000 1200 1400
0
10
20
30
40
50
60
70
80
90
100
load
e f f e i c i e n c y
Relation between effeiciency and load
0 1 2 3 4 5 60.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
P/Po in pu
V r / V s
before compensation
load ability curve
Shunt compensation(c)
series compensation(C)
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Electrical Performance of Transmission LineChapter 2
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2.6 Phenomenon of Corona
2.6.1 1ntroduction :Corona is a phenomenon of ionization of air molecules around the transmission line, and that when
the transmission line of high 220KV and thus voltage distribution is almost equal 30kv/cm It issuitable for ionization of air molecules around the transmission line, and as a result of this ionization
turn the air molecules around the transmission line of the insulating material to the conductive
material.
Figure: (2.21) Phenomenon of Corona
2.6.2 Corona affected by the following factors: state of the surface of the connector (surface roughness disciplinary electric field to irregular
and may occur Coruna)
state of the surrounding gas (humidity - temperature - the type of gas - atmospheric pressure)
shaped electrodes and the distance between them.
2.6.3 Advantage of Corona Reduces transient (i.e. charges includes on the line by light will be dissipated by corona, so corona
acts as safety valve).
2.6.4 Disadvantage of Corona • High power loss in transmission.
• TV & Radio signal interference.
• Hissing noise & conductor vibrate.
• Luminous violet glow around T.L .
• Break down may occur.
• Ozone & oxide of Nitrogen are produced.
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Electrical Performance of Transmission LineChapter 2
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2.6.5 Visual critical voltagež
It is the minimum voltage at which the ionization takes place taking into account that corona
is not visual (complete breakdown of the dielectric).
ž
r
Dr mVrms ln**
2
10*30
6
(2.28)
ž Where
ž mo = Irregularity factor of wire=0.8 for stander conductor .
ž r = Radius of the conductor.
ž D = Distance between the wires.
ž = Reduction factor of break down strength also called air density factor.
(2.29)
Where P = the barometer pressure.
t = the temperature of air in.
r
Dr mVrms ln**
2
10*30
6
(2.30)
8.00 m
mr 310*635.8
799.5 D
9392.0
KvV rms
93.122
In this case does not corona so 122.93Kv > 76.21Kv .
t
P
273
*392.0
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[Mechanical Design of Transmission Line]Chapter 3
03
Chapter 3
Mechanical Design of
Transmission Line
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[Mechanical Design of Transmission Line]Chapter 3
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Mechanical Design of Transmission Line
3.1- Introduction:
Electric power can be transmitted used overhead lines or underground cables as shown in figure 3.1 Electric power in generally transmitted over long distance to substation and as we know that the electric power is
transmitted at high voltages so it need proper insulation. Underground cable is very expensive due to the
above two reasons and it rarely used in transmission systems. Overhead lines are only 15% to 60% costly as
underground cables and therefore it more economically.
In overhead transmission line the mechanical designs and analysis is important to construct good supports to
withstand all expected load and also against the nature powers. The materiel chosen for conductor must be
strong enough to withstand against uncertain whether conditions and forces that will subjected to it.
Conductors also, must have as sufficient electrical characteristic to minimize the power loss and voltage
drop. The overhead lines must provide satisfactory service over long period of time without need to muchmaintenance. Ultimate economy is provided by a good construction since excessive maintenance or short
life can easily more than overbalance a saving in the first cost.
The overhead line must have a proper strength to withstand the stresses imposed on its component parts by
the line itself. These include stresses set up by the tension in conductors at dead points, compression
stresses, transverse loads due to angles in the line, vertical stresses due to the weights of conductors and the
vertical component of conductors tension. It is important to be sure that the conductors are under safe
tension. If the conductors are so much stretched between supports the stress in conductor may reach unsafe
value and conductor may break due to excessive tension.
The poles must have sufficient height and be so located taking into account the geography of the land, as to
provide adequate ground clearance at both maximum loading and maximum temperature.
3.2- Factors affecting mechanical design of overhead lines:
1- Nature of line route.
2- Mechanical loading.
3- Clearance.
4- Conductor.
5- Types of supporting structure.
6- Insulation types.
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[Mechanical Design of Transmission Line]Chapter 3
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(a) (b)
Figure (3.1): (a) Overhead lines. (b) Underground cables
3.3- Nature of line route:
The routes of overhead transmission lines should be selected in order to obtain the most directed route avoid
the building, roads and residential areas. Also, the land of lines route must be selected to make the lines
accessible for maintenance and to reduce the cost of supporting structures. The distance between supporting
towers (span) is in few hundred meters. The factors affecting the length of span are:
Nature of route.
Clearance between conductors.
Excessive tensions under maximum load.
Structures adequate to carry additional loads.
3.4- Mechanical loading:
3.4.1- Stresses:
Mechanical loading refers to the external conditions that produce mechanical stresses in the conductors and
supports. Mechanical loading also includes the weight of the conductors and structures and weight of
equipments at dead point such as cross arms and insulators. Towers supporting and other equipments are
subjected to strains from the tension with which they are strung. When a force is applied against an object it
produces a stress within the object. There are five types of stress:
1-Tensile Stress: Caused by the force acting in opposite directions as shown in figure2.a, for example a
conductors between two towers.
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[Mechanical Design of Transmission Line]Chapter 3
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2-Comprassive Stress: unlike tensile stress is caused by force acting toward the body as shown in figure2.b,
for example the equipments that putting on the poles such as transformer.
3-Shearing Stress: Caused by the forces like compressive stress but not in the straight line that tend to slash body in two as shown in figure2.c, for example bolts attaching a cross-arm to a tower are subjected to a
shearing stress between the cross-arm and the tower.
4-Bending Stress: Caused by the forces acting along a body, for example a tower supporting a corner in the
line and not guyed.
5-Twisting Stress: Caused by line tensions that are not equal on the two sides of a tower, for example when
conductor breaks between supports.
(a) (b) (c)
Figure (3.2): (a) Tension. (b) Compression. (c)Share.
3.4.2- Elasticity:Elasticity is the property of a material that enables it to recover its original shape and size after being
stressed. Importance term called Young modulus is defined as ratio of normal stress to strain and it is
constant for given material up to proportional limit as shown in figure3. Up to certain limit, stress applied to
a material cause's deformation which disappears when the stress is removed. Every material has a stress
limit and beyond this limit (yield point) a certain amount of permanent deformation, this called elastic limit.
When the stress is less than the elastic limit, the deformation is direct proportionally to the stress. When the
stress exceeds the elastic limit the material continue to resist the stress but lost some original characteristic.
If the stress continues the deformation will increase until the material breakdown. The stress caused failure
is the ultimate stress of the material. Some materials have elastic limit and ultimate strength nearly to otherlike glass.
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[Mechanical Design of Transmission Line]Chapter 3
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In design of mechanical structure it is difficult to exact determination of stresses and strengths. The
maximum stress at which structure is design to normally operates is called working stress. The ratio of
working stress to the ultimate strength of the material is the design safety factor. It usual practice to design
for assumed loading conditions and to use this safety factor to make reasonable provision for unusual and
unforeseen conditions and hazards to which the structure subjected to it.
Figure (3.3): Stress-Strain characteristic.
3.4.3- Wind pressure:
To calculate wind pressure on a cylindrical surface like conductor we use Buck formula which given by:
⁄ (3.1)
Where V is the velocity of wind in
⁄
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[Mechanical Design of Transmission Line]Chapter 3
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The pressure on flat surfaces such as cross-arm and tower normal to the direction of wind can be calculated
by using following formula developed by C. F. Marvin:
⁄ (3.2)
In still air the conductor is subjected only to its weight and if the temperature is high the sag in conductorwill increase which decrease the tension. The still air at high temperature is the easiest condition that my
happen in practice. The worst conditions are when the temperature is low which is reduce the sag of the line
and the effect of wind blowing against the conductor.
3.5- Clearance:
As shown in figure4 clearance is required to avoid contact with several objects such as ground, other
conductors on the same structure, buildings, trees, tracks, conductors and structure of another line, the
structure itself and other things can provide a path to ground.
The clearance increase with voltage level there are a minimum required clearances given by NESC for some
voltage levels.
The locations of towers must be chosen to provide sufficient clearance from driveways, fire hydrants, street
traffic, building, etc. Conductors of one line should not less than 1.22m from those of another and
conflicting line. If conductors pass near the tower of another line, providing that they are not attached, they
should not interfere with the climbing space.
3.5.1- Horizontal separation of conductors from each other;
The NESC requires that for supply conductors of the same circuit at voltage up to 8.7kV the minimum
horizontal clearances between the conductors should be 30.5cm as we know that the dielectric strength of air
is 30kV/cm. For higher voltage should be 30.5cm plus 1cm per kV over 8.7kV.The required clearance for
supply conductor of different circuits at voltages up to 8.7kV, the minimum horizontal clearances between
the conductors should be 30.5cm. for voltages between 8.7 and 50kV the clearance should be 30.5cm plus
1cm per kV over 8.7kV and voltages between 50 and 814kV the clearance should be 72.4cm plus 1cm per
kV over 50kV.table1 gives the span length and ground clearance at different voltages.
Figure (3.4): Ground clearance. Table 3.1
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[Mechanical Design of Transmission Line]Chapter 3
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3.6-Type of supporting structure:
The supporting structures for overhead lines are various types of poles and towers called line supports. In
general, the line support should have the following properties:
1- High mechanical strength to withstand the weight of conductors and wind load etc.
2- Light in weight without loss mechanical strength.
3- Cheaper in cost and economical to maintain.
4- Longer life.
5- Easy accessibility of conductors for maintenance.
The types of line support used in transmission and distribution are wood poles, steel poles, concrete poles
and steel towers. The choice of supporting structure for particular case depend upon the line span, cross-
sectional area, line voltage, cost and local conditions.
Wood poles: these are made of seasoned wood and are suitable for lines of moderate cross-sectional area
and of relatively shorter spans up to 50m. This supports are cheap, easily available, provide insulating
properties and therefore, are widely used for distribution purposes in rural areas as economical proposition.
The wood poles generally tend to rot below the ground level causing foundation failure. In order to prevent
this portion of the pole below the ground level is impregnated with preservative compounds like creosote
oil.
The main disadvantages of wood pole are tendency to rot below the ground level, less mechanical strength,
the possibility of combustion and comparatively smaller life 20 to 25 years.
Steel poles: the steel poles are often used as a substitute for wood poles. They have greater mechanical
strength, longer life and longer spans to be used. These poles are used for distribution purpose in cities. This
type of supports needs to be galvanised or painted in order to increase its life. The steel poles are of three
type's rail poles, tubular poles and rolled steel joints.
Concrete poles: The reinforced concrete poles have become very popular as line support in recent years.
They have greater mechanical strength, longer life and permit longer spans than steel poles. Moreover, they
give good outlook, require little maintenance and have good insulating properties. The main difficulty is the
high cost of transport due to their heavy weight.
Steel towers: The wood, steel and concrete poles are often used in distribution system. However, for long
distance transmission at high voltages steel towers are used. Steel towers have greater mechanical strength,
longer life, can withstand most severe weather conditions and permit the use of longer spans. The risk of
interrupted service due to broken or punctured insulation is reduced owing to longer spans. Towers footings
are usually grounded by driving rods into the earth. This minimises the lightning troubles as each tower acts
as a lightning conductor. However, in a moderate additional cost double circuit towers are provided to insure
continuity of supply. In case of breakdown of one circuit the continuity of supply will achieved by the other
circuit.
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[Mechanical Design of Transmission Line]Chapter 3
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3.7- Insulator:
The overhead line conductors should be supported on the poles or towers in such a way that currents from
conductors do not flow to earth through supports. This is achieved by securing line conductors to supports
with the help of insulators. These insulators provide necessary insulation between line conductor and
supports and thus prevent any leakage current from conductors to earth. In general, the insulators should
have the following desirable properties:
1- High mechanical strength in order to withstand conductor load, wind load etc.
2- High electrical resistance of insulator material in order to avoid leakage currents to earth.
3- High relative permittivity of insulator material in order that dielectric strength is high.
4- The insulator material should be non-porous; free from impurities and cracks otherwise the permittivity
will be lowered.
5- High ratio of puncture strength to flashover.
The most commonly used material for insulators of overhead line is porcelain but glass ,steatite and special
composition materials are also used to a limited extent . Porcelain is produced by firing at a high
temperature a mixture of kaolin, feldspar and quartz. It is stronger mechanically than glass, gives less
trouble from leakage and is less affected by changes of temperature. The successful operation of an
overhead line depends to a considerable extent upon the proper choice of insulators. There are several types
of insulators but the most commonly used are pin type, suspension type, strain insulator and shackle
insulator.
1- Pin type insulators: The pin type insulator is secured to the cross-arm on the pole. There is a groove on
the upper end of the insulator as shown in figure 3.5.a for housing the conductor. The conductor passes
through this groove and is bound by the annealed wire of the same material as the conductor. Pin type
insulators are used for transmission and distribution of electric power at voltages up to 33kV. Beyond
operating voltage of 33kV, the pin type insulators become too bulky and hence uneconomical.
2- Suspension type insulators: The cost of pin type insulator increase rapidly as the working voltage is
increased. Therefore, this type of insulator is not economical beyond 33kV. For high voltages (>33kV) it is a
usual practice to use suspension type insulators as shown in figure3.5.b. They consist of number of porcelain
discs connected in series by metal links in the form of a string. The conductor is suspended at the bottom
end of this string while the other end of the string id secured to the cross-arm of the tower. Each unit or disc
is designed for low voltage, say 11kV. The number of discs in series would obviously depend upon the
working voltage. For instant, if the working voltage is 66kV then six discs in series will be provided on the
string.
3- Strain insulators: When there is a dead end of the line or there is corner or sharp curve, the line is
subjected to greater tension. In order to relative the line of excessive tension, strain insulators are used. For
low voltage (<11kV), shackle insulators are used as strain insulators. However, for high voltage
transmission lines, strain insulator consists of an assembly of suspension insulators. The discs of strain
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[Mechanical Design of Transmission Line]Chapter 3
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insulators are used in the vertical plane. When the tension in lines is exceedingly high, as at long river spans,
two or more strings are used in parallel.
4- Shackle insulators:In early days, the shackle insulators were used as strain insulators. But now days,
they are frequently used for low voltage distribution lines. Such insulators can be used either in horizontal
position or in a vertical position. They can be directly fixed to the pole with a bolt or to the cross-arm and
the conductor in the groove is fixed with a soft binding wire.
Figure (3.5): (a) Pin type insulator. (b) Suspension type insulator.
3.8-Sag and Tension analysis:
While erecting an overhead lines, it is important that conductors are under safe tension. If the conductors are
too much stretched between supports in a bid to save conductor material, the stress in the conductor may
reach unsafe value and in certain cases the conductor may break due to excessive tension. In order to permit
safe tension in the conductors, they are not fully stretched but are allowed to have sag. The difference in
level between points of supports and the lowest point on the conductor is called sag.
Conductor sag and tension analysis is an important consideration in overhead transmission and distribution
design. The quality and continuity of electric service supplied depend largely on whether the conductors
have been properly installed. The designer engineer must determine the amount of sag and tension to be
given the wires or cables of a particular line at a given temperature. Tension in the conductors contributes to
the mechanical load on structures at angles in the line and at dead ends. The values of sag and tension for
winter and summer conditions must be known. The factors affecting the sag of a conductor between
supports are:
1- Conductor load per unit length.
2- Span (distance between supports).
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[Mechanical Design of Transmission Line]Chapter 3
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3- Temperature.
4- Conductor tension.
In order to determine the conductor load properly, the factors that need to be taken into account are:1- Weight of conductor itself.
2- Weight of ice or snow clinging to wire.
3- Wind blowing against wire.
From the practical point of view, economic design dictates that conductor sag should be minimum to refrain
from extra pole height, to provide sufficient clearance above ground level and to avoid providing excessive
horizontal spacing between conductors to prevent them swinging together at mid-span. Conductor tension
pulls the conductor up and decrease its sag. At the same time, tension elongates the conductor, from elasticstretching, which tends relieve tension and increase sag. The elastic property of metallic wire is measured by
its modulus of elasticity. The modulus of elasticity is defined as the stress per unit area divided by the
deformation per unit length. Since
⁄ (3.3)
: stress per unit area ( ⁄ )
T: conductor tension(kg)
A: actual cross section of conductor(
)
Elongation e of the conductor due to the tension is
(3.4)
elongation is high when if modulus of elasticity is low. Thus, a small change in the length of conductor
causes large effect on sag and tension of conductor.
Sag and stresses in conductors are dependent on the initial tension put on them when they are clamped in
place, weight of conductors themselves, ice or sleet clinging to them and wind pressure.
Since the stress depends on sag, any span can be used provided the poles or towers are high enough and
strong enough. The matter is merely one of extending the centenary in both directions. But the cost of poles
or towers sharply increases with height and loading. Thus, the problem becomes the balancing of larger
number of lighter and shorter poles or towers against smaller number of heavier and taller ones.
3.8.1-Effect of change in temperature:
Sag and stresses very with temperature according to thermal expansion and contraction of the conductors.
Temperature rise of conductor increase the length of conductor, and hence sag increase and tensiondecreases. A temperature fall cause opposite effect. Maximum stress occurs at the lowest temperature, when
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[Mechanical Design of Transmission Line]Chapter 3
3
the line has contracted. If conductor stress is constant while the temperature is changes, the change in length
of the conductor is
(3.5)
(3.6)
Where
to: initial temperature
lo: conductor length at initial temperature to
l1: conductor length at t1
α: coefficient of linear expansion of conductor per degree centigrade
change in temperature in degree centigrade
change in conductor length in meter :
if the temperature is constant while conductor stress change (i.e. loading) the change in length of the
conductor is
(3.7)
(3.8)
Where
: initial tension of conductor(kg)
: change in conductor tension(kg)
M : modulus of elasticity of conductor(kg.m)
A : actual metal cross section of conductor( )
3.8.2 Calculation of line sag and tension
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[Mechanical Design of Transmission Line]Chapter 3
A conductor suspended freely from tow supports, which are at the same level and span L unit length apart as
shown in figure8, takes the form of a catenary curve providing the conductor is perfectly flexible and its
weight is uniformly distributed along its length. If the conductor is tightly stretched (i.e. when sag d is very
small compared to span L), the resultant curve can be considered a parabola. If the conductor sag is less than
6 percent of its span length, the error in sag computed by the parabolic equations is less than 0.5 percent. Ifthe conductors sag is less than 10 percent of the span, the error is about 2 percent.
Figure (3.6): Conductor suspended between supports at same levele
3.8.3-Supports at same level
Catenary method:
Figure 3.6 shows a span of conductor with two support at the same level and separated by a horizontal
distance L. Let O be the lowest point on the catenary curve l be the length of conductor between tow
supports. Let w be the weight of the conductor per unit length, T be the tension of the conductor at any point
P in the direction of the curve and H be the tension at origin O. Further, let s be the length of the curve
between O and P, so that the weight of the portion s is :
Or approximately
The total tension in the conductor at any point x is
The total tension in the conductor at the support is when ,is
(3.9)
(3.10)
(3.11)
(3.12)
(3.13)
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[Mechanical Design of Transmission Line]Chapter 3
The sag of the conductor for a span of length L between two supports at the same level
Or approximately
3.8.4 Effect of wind:
We assume that wind blows uniformly and horizontally across the projected area of the conductor. Figure 3.7 showsthe force of wind on conductor
Figure (3.7): Wind force on conductor.
The projected area per meter length of the conductor is S=Al (3.15)WhereS: projected area of conductor in square meterA: cross-sectional area of conductor in square meterl : length of conductor in meterFor 1-meter length of conductor
⁄
The horizontal force exerted on the line as a result of the pressure of wind is
For 1-meter length of conductor
⁄
Wheredc: diameter of conductor (cm)
F : horizontal force due to wind pressure exerted on line ( ⁄ )
P : wind pressure ( ⁄ )The effective load acting on the conductor is
√ ⁄
Note: we neglect the effect of ice because it rarely falls here in Jazan region.
(3.16)
(3.17)
(3.18)
(3.19)
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[Mechanical Design of Transmission Line]Chapter 3
0
3.8.5 Calculation of line sag and tension of our project :
The weight of the cable ,
W=1302 kg/km = 1.302 kg/m
Assume the length between towers :
l =250 m
The tension of cable :
T=111.9 KN, 1Kg=9.807 N → =0.009807 kN
T=11410.21 N
Sag :
Fig 3.8 conductor suspended between
supports at same level
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[Mechanical Design of Transmission Line]Chapter 3
Use factor safety =3
Effect of wind :
o Wind Pressure (Pw)= Ib/ft2
We found the average speed of wind in jazan region is
o Vm=18 Km/hr =11.78 mile/hr .
So , the wind pressure is equal :
o Pw= Ib/ft
2 =
kg/m
2
Wind load (Ww) per unit length =
o Ww= (
) =
o Where
Dc : Conductor diameter , t: thickness of sand
Assume t=0.2cm =0.0065 ft
o
( )
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[Power Flow Analysis]Chapter 4
4
Chapter 4
POWER FLOW
ANALYSIS
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[Power Flow Analysis]Chapter 4
Power Flow Analysis
4.1 Introduction
Power flow studies are performed to determine voltages, active and reactive power etc. an various
points in the network for different operating conditions subject to the constraints on generatorcapacities and specified net interchange between operating systems and several other restraints.
Power flow or load flow solution is essential for continuous evaluation of the performance of the
power systems so that suitable control measures can be taken in case of necessity. In practice it will
be required to carry out numerous power flow solutions under avariety of conditions.
For the purpose of power flow studies a single phase representation of the power network
is used, since the system is generally balanced.
4.2 Necessity for Power Flow Studies
Power flow studies are undertaken for various reasons, some of which are the following:1. The line flows
2. The bus voltages and system voltage profile
3.
The effect of change in configuration and incorporating new circuits on system
4. loading
5.
The effect of temporary loss of transmission capacity and (or) generation on system
6. loading and accompanied effects.
7.
The effect of in-phase and quadrate boost voltages on system loading
8. Economic system operation
9. System loss minimization
10. Transformer tap setting for economic operation
11.
Possible improvements to an existing system by change of conductor sizes and12.
system voltages.
4.3 Bus Classification
Each bus in the system has four variables: voltage magnitude, voltage angle, real power and reactive
power. During the operation of the power system, each bus has two known variables and two
unknowns. Generally, the bus must be classified as one of the following bus types:
4.3.1 Slack or Swing Bus
This bus is considered as the reference bus. It must be connected to a generator of high rating relative
to the other generators. During the operation, the voltage of this bus is always specified and remains
constant in magnitude and angle. In addition to the generation assigned to it according to economic
operation, this bus is responsible for supplying the losses of the system.
4.3.2 Generator or Voltage Controlled Bus
During the operation the voltage magnitude at this the bus is kept constant. Also, the active power
supplied is kept constant at the value that satisfies the economic operation of the system. Most
probably, this bus is connected to a generator where the voltage is controlled using the excitation and
the power is controlled using the prime mover control (as you have studied in the last experiment).
Sometimes, this bus is connected to a VAR device where the voltage can be controlled by varying the
value of the injected VAR to the bus.
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[Power Flow Analysis]Chapter 4
4.3.3 Load Bus
This bus is not connected to a generator so that neither its voltage nor its real power can be
controlled. On the other hand, the load connected to this bus will change the active and reactive
power at the bus in a random manner. To solve the load flow problem we have to assume the
complex power value (real and reactive) at this bus.
Bus classification is summarized in Table (4.1)
Table (4.1): Bus classification 4.4 Load Flow Equations:Assuming a system having n buses, the injected current to the bus (node) k can be expressed as:
N
n
nknk V Y I 1
(4.1)
Where Yknis the proper element in the bus admittance matrix YBus
and,
||Y Y , ||V V (4.2)The complex power at bus k ( k = 1,2, ..,n) is given as:
k k k k k jQ P I V S **
(4.3)
Where, Pk
and Qk
are the active and reactive power injection at bus k respectively. Thus, at each bus
we have two equations and four variables ( P, Q, δ, V ). Note that Y's and θ's are known from
network data. Actually, at each bus we have to specify two variables and solve for the remaining two
unknowns.
4.5 Techniques of Solution
Because of the nonlinearity and the difficulty involved in the analytical expressions for the above
power flow equations, numerical iterative techniques must be used such as:
1. Gauss-Seidel method (G-S).
2. Newton-Raphson method (N-R).
The first method (G-S) is simpler but the second (N-R) is reported to have better convergence
characteristics and is faster than (G-S) method.
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[Power Flow Analysis]Chapter 4
4.5.1 Gauss - Seidel Iterative Method of Load Flow Solution
In this method, voltages at all buses except at the slack bus are assumed. The voltage at the slack bus
is specified and remains fixed at that value. The (n-1) bus voltage relations.
n
ik k
k ik
n
ii
ii
i V Y V
jQ P
Y V
1
*
1 (4.4)
i=1,2,3,….,n; i≠ slack bus .
In order to accelerate the convergence, allnewly-computed values of bus voltages are substituted in
eqn. (4.4). Successively the busvoltage equation of the (m + 1)th iteration may then be written as :
n
ik
n
ik
m
k ik
m
k ik m
n
ii
ii
m
i V Y V Y V
jQ P
Y V
1
)()1(
)*(
)1( 1 (4.5)
The solution procedure is to:
1. Initialize the bus voltages. For load busses, use V = 1 + j0. For generator busses (including
the swing bus),
2. One-by-one, update the individual bus voltages using equation (4.4)
For PV busses, update the voltage angle, while holding the voltage magnitude constant at the
specified value. Do not update the swing bus.
3. Check the mismatch P and Q at each bus. If all are within tolerance (typical tolerance is
0.00001 pu), a solution has been found. Otherwise, return to Step 2.
4.5.2The Newton-Raphson Power-Flow Solution
Newton-Raphson (NR) method is more efficient and practical for large power systems. Main
advantage of this method is that the number of iterations required to obtain a solution is independentof the size of the problem and computationally it is very fast. Here load flow problem is formulated
in polar form.
N
n
nknk k k V Y V jQ P 1
* (4.6)
)cos(||||||1
N
n
knk nnknk k V Y V P (4.7)
)sin(||||||1
N
n
knk nnknk k V Y V Q (4.8)
Update the above equations at (k=n ) :
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[Power Flow Analysis]Chapter 4
)cos(||||||||||1
2
N
n
knk nknnk kk k k Y V V GV P (4.9)
N
n
knk nknnk kk k k Y V V BV Q1
2 sin(|||||||||| (4.10)
Equations (4.9) and (4.10) constitute a set of nonlinear algebraic equations in terms of the
independent variables, voltage magnitude in per unit and phase angles in radians, we can easily
observe that two equations for each load bus given by eqn. (4.9) and (4.10) and one equation for each
voltage controlled bus, given by eqn. (4.9) Expanding eqns. (4.9) and (4.10) in Taylor-series and
neglecting higher-order terms. We obtain,
Normally, only 3 to 5 iterations are required to solve the loadflow problem, regardless of system size.
Newton-Raphson is the most commonly used loadflow solution technique.
(4.11)
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[Power Flow Analysis]Chapter 4
45
4.6 Load flow by using Matlab program
REPORT OF POWER FLOW CALCULATIONS
Swing Bus : BUS 1
Total Real Power Losses : 1.27927
Total Reactive Power Losses: -1.87689
Table (4.2): load flow of the network
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[Power Flow Analysis]Chapter 4
4
4.7 Load flow by using Power world Simulator Program :
Figure (4.1) equivalent electrical network of Jazan
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[Fault System Analysis]Chapter 5
25
Chapter 5
FAULT SYSTEMANALYSIS
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[Fault System Analysis]Chapter 5
2
Fault System Analysis
5.1Introduction :
Fault studies form an important part of power system analysis and the problem consists of
determining bus voltage and line current during faults. Fault studies are used for proper choice of
circuit breakers and protective relaying.
Faults divided in to two section :
1- Symmetrical Short circuit (3-phace)
2- Unsymmetrical Short Circuit
-Line – Ground (L-G)
-Line To Line (L-L)
-Doable Line With Ground (2L-G)
5.2 Objectives of faults analysis :
1- Calculate rating of circuit breaker (C.B) to isolate the fault , therefore to with stand the
equipment from damage.
2- To get acceptable setting of relay .
5.3 Types of common faults :
There are four common faults may cause balanced and / or unbalanced operation conditions. These
faults with their associated relative occurrence frequencies are listed in Table 5.1
Table 5.1: fault type and their frequencies
Relative Frequency Fault Type
70% Single line to ground (L-G)
15% Line to Line (L-L)
10% Doable Line With Ground (2L-G)
5% Three Phase (3P)
100% Total
5.4 Symmetrical Short circuit (3-phase) :
This section is devoted to the analysis of symmetrical three-phase fault or balanced fault. This type
of fault can be defined as the simultaneous short circuit across all the three phases. This type of fault
occurs infrequently,. The three phase fault information is used to select and set phase relays.
So , the conditions of three phase balanced system :
1) Voltage and current magnitude are equal
2) Phase difference between each phase equal 120o .
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[Fault System Analysis]Chapter 5
2
Another important point is that the circuit breakers rated MVA breaking capacity is based on three
phase fault MVA. In fact high precision is not necessary when calculating the three phase fault level
because circuit breakers are manufactured in standard sizes, e.g., 250, 500, 750 MVA etc.
Generally for three phase fault calculation, following assumptions are made:
1. The emfs of all generators are 01 pu. This assumption simplify the problem .
2. Charging capacitances of the transmission line are ignored.
3. Shunt elements in the transformer model are neglected.
5.5 Unsymmetrical Short Circuit:
In a balanced system, analysis can be done on a single phase basis. The knowledge of voltage and
current in one phase is sufficient to determine the voltages and current in other two phases. Real and
reactive powers are three times the corresponding per phase values. When the system is unbalanced,the voltages, currents and the phase impedances are in general unequal. Unbalanced system operation
can result due to unsymmetrical fault, e.g., line to line fault, double line to ground fault or single line
to ground fault. Unbalanced operation may also result when loads are unbalanced. Such an
unbalanced operation can be analyzed through symmetrical components where the unbalanced three
phase voltages and currents are transformed into three sets of balanced voltages and currents called
symmetrical components.
Condition of un-balanced three phase system :
- Magnitude V/I are not equal , but phase difference equal 120o
-
Magnitude V/I are equal , but phase difference not equal 120o
-
Both V/I magnitude and phase angle difference are not equal .
Causes of system un-balanced :
- Unbalance load .
- Open one of each phase .
- Unbalance short circuit .
5.5.1 Symmetrical Components:
A system of three unbalanced phasors can be resolved in the following three symmetrical
components: Positive Sequence: A balanced three-phase system with the same phase sequence as the
original sequence.
Negative sequence: A balanced three-phase system with the opposite phase sequence as theoriginal sequence.
Zero Sequence: Three phasors that are equal in magnitude and phase.
Fig. 5.1 depicts a set of three unbalanced phasors that are resolved into the three sequence
components mentioned above. In this the original set of three phasors are denoted by V a, V b and V c,
while their positive, negative and zero sequence components are denoted by the subscripts 1, 2 and 0
respectively. This implies that the positive, negative and zero sequence components of phase-a are
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[Fault System Analysis]Chapter 5
22
denoted by V a1, V a2 and V a0 respectively. Note that just like the voltage phasors given in Fig. 5.1 we
can also resolve three unbalanced current phasors into three symmetrical components.
Figure. (5.1) Representation of (a) an unbalanced network, its (b) positive sequence,
(c) negative sequence and (d) zero sequence.
After the introduction of the positive , negative and zero sequence components, any three-phase
quantities , balanced or unbalanced , can be exoressed in trems of these components instance , for
phase voltage ,
210 aaaa V V V V
21021
2
0 bbbaaab V V V aV V aV V (5.1)
2102
2
10 cccaaac V V V V aaV V V
Where "a" is define as a vector having 1 as its magnitude and 120o as its phase angle.
Namely, oa 1201 . Therefore ,
13601
1201
23
2
o
o
aaa
aaa (5.2)
Clearly , the sum of them equal zero .
01 2 aa (5.3)
Then , the three phase voltages can be re-written in a matrix notation
2
1
0
2
1
0
2
2][
1
1
111
a
a
a
a
a
a
c
b
a
V
V
V
T
V
V
V
aa
aa
V
V
V
The symmetrical component transformation matrix is then given by
c
b
a
a
a
a
V
V
V
aa
aa
V
V
V
2
2
2
1
0
1
1
111
3
1
(5.4)
(5.5)
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[Fault System Analysis]Chapter 5
2
If all of these quantities for current , then
2
1
0
][
a
a
a
c
b
a
I
I
I
T
I
I
I
(5.6)
The fault analysis is summarized as follows :
5.5.2 Single Line to ground (L-G):
The general representation of a single line-to-ground fault is shown in Figure 5.2 where F is the
fault point with impedances Zf . Figure 5.3 shows the sequences network diagram. Phase 'a' is
usually assumed to be the faulted phase, this is for simplicity in the fault analysis calculations.
a
c
b
+
Vaf
-
F
Iaf I bf = 0 Icf = 0
n
Zf
Figure (5.2) General representation of a single line-to-ground fault.
Xo
X2
X1
Ia1
Ia2
Ia0
Va2
Va1
Va0
3Zf
o01
Figure (5.3) Sequence network diagram of a single line-to-ground fault.
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[Fault System Analysis]Chapter 5
2
Since the zero-, positive-, and negative-sequence currents are equals as it can be observed in
Figure 5.2. Therefore,
0 1 2
0 1 2
1.0 0
3a a a
f
I I I Z Z Z Z
(5.7)
The voltage at faulted phase a can be obtained as :
13af f aV Z I (5.8)
With the results obtained for sequence currents, the sequence voltages can be obtained as :
0 0 0
1 1 1
2 2 2
1.0
a a
a a
a a
V Z I
V Z I
V Z I
(5.9)
5.5.3 Line-to-line fault : The general representation of a line-to-line fault is shown in Figure 5.3 where F is the fault point with
impedances Zf . Figure 5.4 shows the sequences network diagram. Phase b and c are usually assumed
to be the faulted phases; this is for simplicity in the fault analysis calculations .
a
c
b
F
Iaf = 0 I bf Icf = -I bf
Zf Figure (5.4) General representation of a single line-to-line fault.
X1
Ia1o01
Zf
Va1Ia2
Figure (5.5) Sequence network diagram of a line-to-line fault. From Figure 5.4 it can be noticed that
0af I
bf cf I I (5.10)
bc f bf V Z I
Note that , zero sequence not excite , positive and negative networks in parallel ,without zerosequence .
Therefore, the sequence currents can be obtained as :
0 0a I (5.11)
1 2
1 2
1.0 0a a
f
I I Z Z Z
(5.12)
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[Fault System Analysis]Chapter 5
2
The sequence voltages can be found as :
0
1 1 1
2 2 2 2 1
0
1.0-
a
a a
a a a
V
V Z I
V Z I Z I
(5.13)
5.5.4
Double Line-to-Ground FaultThe general representation of a double line-to-ground fault is shown in Figure 3.14 where F is
the fault point with impedances Zf and the impedance from line to ground Zg . Figure 3.15 shows the
sequences network diagram. Phase b and c are assumed to be the faulted phases, this is for simplicity
in the fault analysis calculations.
a
c
b
F
Iaf = 0 I bf Icf
n
Zf Zf
Zg I bf +Icf
N
Figure (5.6) General representation of a double line-to-ground fault.
X1
Ia1o01
3Zf
X2
Ia2 Ia0
Figure (5.7) Sequence network diagram of a double line-to-ground fault.
To get positive sequence ' 1a I ' ;
X1
Ia1 Z1
Figure (5.8) : Simple circuit represented of a double line-to-ground fault.
Where :
o f
o f
jx Z jx
jx Z jx Z
3
)3(
2
2
1
11
1
01
Z jx I
o
a
(5.14)
The negative and zero sequence current are equal :
f
f
aa Z jx jx
Z jx I I
3
3
02
0
12 (5.15)
f
aa Z jx jx
jx I I 3
02
210 (5.16)
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[Fault System Analysis]Chapter 5
2
5.6 Short circuit calculations :
Table (5.2) show the different values of bus voltage with different locations of short circuit.
Table (5 .2)
ISC bus 132 kv
V(13) medical
city(pu)
V(6) kodmi (pu)
1 o87.090453.0 o24.2902386.0
2 o40.487032.0 o
51.786823.0
3 o19.2068438.0 o30.2368274.0
4 o16.3049239.0 o27.3349121.0
5 o10.741285.0 o21.1041186.0
6 o00 o00
7 o29.1437263.0 o40.1737173.0
8 o22.2140228.0 o33.2440132.0
9 o36.55595.0 o47.855823.0
10 o72.2819322.0 o82.3119276.0
11 o48.1762432.0 o59.2062282.0
12-14 o57.1558801.0 o67.1858660.0
13 o00 o26.1250609.0
Where the impedance between bus 6 and bus 13 is equal to
In polar form :
o76.67
To get the short circuit current we apply in this formal :
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[Fault System Analysis]Chapter 5
6
the short circuit currents with different values of bus voltage show in table (5.3)
Table (5.3)short circuit currents
puo42.254387.0 Ish1
puo73.3042217.0 Ish 2
puo4.463319.0 Ish 3
puo
5.562388.0 Ish 4
puo42.3320.0 Ish 5
pu0Ish 6
puo64.401807.0 Ish 7
puo75.222133.0 Ish 8
puo55.312714.0 Ish 9
puo03.550934.0 Ish 10
puo81.433028.0 Ish 11
puo14.422678.0 Ish 12
puo5.55524.4 Ish 13
√
√
√
C.B rating standard :250,500,750 MVA
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[Power System Stability]Chapter 6
16
Chapter 6
Power System Stability
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[Power System Stability]Chapter 6
1
Power System Stability
6.1 Introduction
An electrical power system consists of many individual elements connected to gether to form a large,
complex system capable of generating, transmitting and distributing electrical energy over a largegeographical area. Due to these interconnections of elements, a large variety of dynamic interactions
are possible to done which may effect on the system. The stability problem involves the study of the
electromechanical oscillations inherent in power systems . Power systems exhibit various modes of
oscillation due to interactions among system components. Power systems usually have two distinct
forms of oscillations.
1.
Local modes are associated with the swinging of units at a generating station with respect to
the rest of the power system. The term local is used because the oscillations are localized at
one station or a small part of the power system. Typical local-mode frequency range from 0.8-2.0 Hz.
2. Inter-area modes are associated with the swinging of many machines in one part of the system
against machines in other parts. They are caused by two or more groups of closely couple
machines being interconnected by weak ties. Typically have a frequency in the range from
0.1-0.8 Hz.
Undamped oscillations once started often grow in magnitude over the span of many seconds.
Sustained oscillations in the power system are undesirable for many reasons. They can lead to fatigueof machine shafts, cause excessive wear of mechanic calactuators of machine controllers and also
make system operation more difficult. It is therefore desirable that oscillations are well damped. So
PSS is necessary to provide appropriate damping of undesirable oscillations caused by disturbances.
6.2 Power System Stabilizers
The basic function of a power system stabilizer is to extend the stability limits by adding damping to
generator rotor oscillations by controlling its excitation using auxiliary stabilizing signal(s). To
provide damping, the stabilizer must produce a component of electric torque, which is in phase with
rotor speed deviations. The oscillations of concern typically occur in the frequency range of
approximately 0.1 to2.0 Hz, and insufficient damping of these oscillations may limit the ability totransmit the power. The block diagram used in industry is shown in Figure (6.1). It consists of a
washout circuit, phase compensator (lead-lag circuit), stabilizer gain and limiter .
Stabilizer gain Washout Lead-Lag Lead-Lag
STAB
K
W
W
sT
sT
12
1
1
1
sT
sT
4
3
1
1
sT
sT
maxu
minu
u
Figure (6.1) Block diagram of PSS
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[Power System Stability]Chapter 6
1
The phase compensation block provides the appropriate phase lead characteristic to compensate for
the phase lag between the exciter input and generator electricaltorque. The required phase lead can be
obtained by choosing the values of timeconstants 1 4 . T ,....,T
The signal washout block serves as a high pass filter, with the time constant high enough to allow
signals associated with oscillations in speed to pass as it. Without it, steady changes in speed wouldmodify the terminal voltage. It allows the PSS torespond only for a change in the speed. From the
viewpoint of the washout function,the value of is not critical and may be in the range of 1 to 20
seconds.
The stabilizer gain K STAB determines the amount of damping introduced by PSS. Ideally the gain
should be set at a value corresponding to maximum damping.However, in practice the gain is set to a
value that results in satisfactory damping ofthe critical system modes without compromising the
stability of other modes.
In order to restrict the level of generator terminal voltage fluctuation during transientconditions,
limits are imposed on PSS outputs.
6.2.1 PSS gain, time constant and block diagram
The value of the washout time constantW T is kept at 10 second, the values of time constants
2T
and4T are fixed at a reasonable value of 0.05 second. The stabilizer gain K and time constants
1T ,
and3T are remained to be determined.
K=25.3696;1T =0.4684;
3T =0.7428
Range of T2, T4 Form (0.06 to 2), gain K from (1 to 50)TA , K A are gain time constant of excitation system =
TA= 0.05 second , K A =400 .
Figure.(6.2). Block diagram of PSS with excitation system.
-
+
+
(
) (
)
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[Power System Stability]Chapter 6
1
6.2.2 Block diagram of PSS in Matlab Simulink Program :
Figure (6.3) show the block diagram of PSS stabilizerin Matlab program .
Figure (6.3): the block diagram of PSS stabilizer
Result :The result show the change of delta (δ) of system stability with and without stabilizer ( PSS ) per
second as in shown in figure (6.4)
Figure( 6.4) : change of omega(∆ω) with time per second.
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[Distribution System for Residential area]Chapter 7
[56]
Chapter 7
Distribution System
for Residential area
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[Distribution System for Residential area]Chapter 7
[55]
Distribution System for Residential area 7.1 Introduction :
Wires carrying current always have inherent resistance, or impedance, to current flow. Voltage drop
is defined as the amount of voltage loss that occurs through all or part of a circuit due to impedance.
Excessive voltage drop in a circuit can cause lights to flicker or burn dimly, heaters to heat poorly,
and motors to run hotter than normal and burn out. This condition causes the load to work harder
with less voltage pushing the current.
The National Electrical Code recommends limiting the voltage drop from the breaker box to the
farthest outlet for power, heating, or lighting to 3 percent of the circuit voltage. This is done by
selecting the right size of wire and is covered in more detail under
7.2 Causes of voltage drops
Voltage drop is caused by resistance in the conductor or connections leading to the electrical load.
There are many causes of resistance in the conductor path.:
7.2.1 There are four fundamental causes of voltage drop
i. Material - Copper is a better conductor than aluminum and will have less voltage drop than
aluminum for a given length and wire size. Copper is a better conductor than aluminum and
will have less voltage drop than aluminum for a given length and wire size.
ii. Wire Size - Larger wire sizes (diameter) will have less voltage drop than smaller wire sizes
(diameters) of the same length.
iii. Wire Length - Shorter wires will have less voltage drop than longer wires for the same wire
size (diameter).
iv. Current Being Carried - Voltage drop increases on a wire with an increase in the current
flowing through the wire.
Figure (7.1) Residential area
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[Distribution System for Residential area]Chapter 7
[5]
7.3 Application Of Capacitors to Distribution Systems :
7.3.1 Basic Definitions
Capacitor element : an indivisible part of a capacitor consisting of electrodes separated by adielectric material.
Capacitor unit : an assembly of one or more capacitor elements in a single container with terminals
brought out.
Capacitor segment : a single-phase group of capacitor units with protection and control system.
Capacitor module: a three-phase group of capacitor segments Capacitor bank: a total assembly of
capacitor modules electrically connected to each other.
7.4 Power Factor Improvement:
At a casual look a capacitor seems to be a very simple and unsophisticated apparatus, i.e., two metal plates separated by a dielectric insulating material. It has no moving parts, but instead functions by
being acted upon by electric stress. In reality, however, a power capacitor is a highly technical and
complex device in that very thin dielectric materials and high electric stresses are involved coupled
With highly sophisticated processing techniques. In the past, most power capacitors were
constructed with two sheets of pure aluminum foil separated by three or more layers of chemically
impregnated Kraft paper. Power capacitors have been improved tremendously over the last 30 years
or so, partly due to improvements in the dielectric materials and their more efficient utilization and
partly due to improvements in the processing techniques involved. Capacitor sites have increased
from the: 15-25 kvar range to the 200-300 kinar range (capacitor banks are usually supplied in sizes
ranging from 300 to 1800 kvaf) nowadays power capacitors are much more efficient than those of
30 years Ago and are available to the electric utilities at a much lower cost per kilo var. In general,capacitors are getting more attention today than ever before, partly due to a new dimension added in
the analysis: change out economics. Under certain circumstances, even replacement of older
capacitors can be justified on the basis of lower-loss evaluations of the modem capacitor design.
Capacitor technology has evolved to extremely low loss designs employing the all-film concept; as
a Result, the utilities can make economic loss evaluations in choosing between the presently existing
capacitor technologies.
7.4.1 Fixed Versus Switched Capacitors:
Shunt capacitors applied to distribution systems are generally located on the distribution lines or inthe substations. The distribution capacitors may be in pole-mounted racks, pad-mounted banks, or
submersible installations. The distribution banks often include three to nine capacitor units
connected in three-phase grounded wye, ungrounded wye, or in delta configuration. The distribution
capacitors are intended for local power factor correction by supplying reactive power and
minimizing the system losses. The distribution capacitors can be fixed or switched depending on the
load conditions.
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[Distribution System for Residential area]Chapter 7
[5]
The following guidelines apply:
i. Fixed capacitors for minimum load condition.
ii. Switched capacitors for load levels above the minimum load and up to the peak load.
Figure 7.2 shows the reactive power requirements of a distribution system are shown for a
period of 24 hours. Such
base load and peak load conditions are common in most utilities. Usually, the fixed capacitors
satisfy the reactive power requirements for the base load and the switched capacitors compensate
the inductive kVAR requirements of the peak load.
7.4.1.1Sizing and Location of Capacitors
To obtain the best results, shunt capacitors should be located where they produce maximum loss
reduction, provide better
voltage profile, and are close to the load. When this is not practical, the following approaches can be
used.
i. For uniformly distributed loads, the capacitor can be placed at two thirds of the distance
from the substation.
ii. For uniformly decreasing distributed loads, the capacitor can be placed at half the distance
from the substation.iii. For maximum voltage rise, the capacitor should be placed near the load.
Usually, the capacitor banks are placed at the location of minimum power factor by measuring the
voltage, current, kW, kVAR, and kVA on the feeder to determine the maximum and minimum load
conditions. Many utilities prefer a power factor of 0.95. The peaks and valleys in the kVAR demand
curve make it difficult to use a single fixed capacitor bank to correct the power factor to the desired
level. If a unity power factor is achieved during the peak load, then there would be leading kVAR
on the line during off-peak condition, resulting in an over-corrected condition. Over-correction of
power factor can produce excess loss in the system, similar to the lagging power factor condition.
Overvoltage condition may occur during leading power factor condition causing damage to the
equipment. Therefore, a leading power factor is not an advantageous condition. In order to handlesuch conditions, fixed capacitors are used to supply the constant kVAR requirements and switched
Figure (7.2 )Distribution curve showing the base reactive power requirement (for fixed
capacitors) and peak kVAR needs (for switched capacitors) switched capacitors).
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[Distribution System for Residential area]Chapter 7
[5]
capacitors are used for supplying the kVAR for the peak load conditions. Specifically, this will
prevent over-correction of the power factor. Figure 7.2 shows the selection approach for the fixed
and switched capacitors. The capacitive kVAR required to correct the given power factor to the
desired level can be calculated as shown below.
=Power factor angle of the given load= Desired power factor angle
kW = Three-phase real power
kVAR = from shunt capacitors,
=
The above relation can be expressed in the form of a chart To determine the needed capacitive
reactive power, select the multiplying factor that corresponds tothe present power factor and the desi
red power factor. Thenmultiply this factor by the kW load of the system. Select a
kVAR bank close to the required kVAR.
7.4.1.2 Effect of Shunt Capacitors on Radial Feeders :
Fixed capacitors can be used to improve the power factor on radial feeders [3]. The capacitors can
be located at the source or at the load. In a radial system, the capacitor can be located very close to
the load as shown in (Figure 7.3) . The voltage profile during light load on the radial system without
and with shunt capacitors is shown in (Figure 7.4) . The voltage drop effects are dominant in the
radial system when shunt capacitors are not present. The voltage rise effects are seen when the
capacitor is present and during light load conditions. The voltage profile along the heavily loaded
condition is shown :
Figure( 7.4) Voltage profile along the feeder during light load.
Figure ( 7.3) Example radial
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[Distribution System for Residential area]Chapter 7
[7]
in (Figure 7.5) The voltage profile is within allowed limits with the shunt capacitors. Therefore,
there is always a need to find the optimum location for the installation of fixed shunt capacitors.
7.4.2 Fixed and Switched Capacitor Applications :
To improve the voltage profile, fixed capacitors are used in high voltage circuits at a point two
thirds of the distance from the source. The shunt capacitor size will be determined by the voltage
rise during light load conditions. If too much shunt compensation is connected, the losses in the
system will increase during light load conditions.
After adding the fixed capacitors according to the above guideline, switched capacitors are added to
about two thirds of the peak load reactive power requirements. The switched capacitors should be
added to a location from a voltage correction point of view, but generally in the last one third of the
circuit, close to the load. The switching can be performed using remote control mechanisms.
When shunt capacitors are used for improving the voltage profile, the fixed capacitors should not
increase the voltage at light load above the allowed values. Additional capacitors can be switched toimprove the power factor to 0.95 at rated load .
7.5 System Benefits :
Using shunt capacitors to supply the leading currents required by the load relieves the generator
from supplying that part of
the inductive current. The system benefits due to the application of shunt capacitors include
i. Reactive power support.
ii. Voltage profile improvements.
iii. Line and transformer loss reductions.iv. Release of power system capacity.
v. Savings due to increased energy loss.
These benefits apply for both distribution and transmission systems.
7.5.1 Effects Of Series And Shunt CapacitorsAs mentioned earlier, the fundamental function of capacitors, whether they are series or shunt,
installed as a single unit or as a bank, is to regulate the voltage and reactive power flows at the point
where they are installed. The shunt capacitor does it by changing the power factor of the load,
whereas the series capacitor docs it by directly offsetting the inductive reactance of the circuit to
which it is applied.
Figure (7.5 )Voltage profile along the feeder during heavy load.
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[Distribution System for Residential area]Chapter 7
[]
7.5.2 series CapacitorsSeries capacitors, capacitors connected in series with lines, have been used to a very limited extent
on distribution circuits due to being a more specialized type of apparatus with a limited range of
application. Also, because of the special problems associated with each application, there is a
requirement for a large amount of complex engineering investigation. Therefore, in general, utilities
are reluctant to install series capacitors, especially of small sizes.As shown in (Fig. 7.6) , a series capacitor compensates for inductive reactance. In other words, a
series capacitor is a negative (capacitive) reactance in series with the circuit's positive (inductive)
reactance with the effect of compensating for part or all of it. Therefore, the primary effect of the
series capacitor is to minimize, or even suppress, the voltage drop caused by the inductive reactance
in the circuit. At times, a series capacitor can even be considered as a voltage regulator that provides
for a voltage boost which is proportional to the magnitude and power factor of the through current.
Therefore, a series capacitor provides for. a voltage rise which increases automatically and
instantaneously as the load grows. Also, a series Capacitor produces more net voltage rise than a
shunt capacitor at lower power factors, .which creates more voltage drop. However, a series
capacitor betters the
system power factor much less than a shunt capacitor and has little effect on the source current.Consider the feeder circuit and its voltage-phasor diagram as shown in (Fig 7.6) a and c.
The voltage drop through the feeder can be expressed approximately as
Where
R = resistance of feeder circuit
= inductive reactance of feeder circuit
= receiving- end power factor
= sine of the receiving -end power-factor angleAs can be observed from the phasor diagram, the magnitude of the second term in Eq. (7.6) is much
larger than the first. The difference gets to be much larger when the power factor is smaller and theratio of is small.
However, when a series capacitor Is applied , as shown in (Fig. 7.6) b and d,
Figure (7.6) Voltage-phasor diagrams for a feeder circuit of lagging power factor: (a) , (c)without (b) and (d )with series capacitors.
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[Distribution System for Residential area]Chapter 7
[]
The resultant lower voltage drop can be calculated as
Where = capacitive reactance of the series capacitor.
7.5.3 Shunt Capacitors
Shunt capacitors, i.e., capacitors connected in parallel with lines, are used extensively in distributionsystems. Shunt capacitors supply the type of reactive power or current to counteract the out-of-phase
component of current required by an inductive load. In a sense, shunt capacitors modify the
characteristic of an inductive load by drawing a leading current which counteracts some or all of the
lagging component of the inductive load current at the point of installation .Therefore a shunt
capacitor has the same effect as an overexcited synchronous condenser, generator, or motor.
As shown in (Fig. 7.7) by the application of shunt capacitor to a feeder, the magnitude of the source
current can be reduced, the power factor can be improved and consequently the voltage drop
between the sending end and the load is also reduced.
However, shunt capacitors do not affect current or power factor beyond their point of application.
( Fig. 7.7 ) a and c show the single-line diagram of a line and the voltage-phasor diagram before the
addition of the shunt capacitor and (Fig. 7.7) b and d show them after the addition. Voltage drop infeeders, or in short transmission lines, with lagging power factor can be approximated as
where
R = total resistance of feeder circuit, Ω
= total inductive reactance of feeder circuit. Ω
= real power (or in-phase) component of current, A
= reactive (or out-of-phase) component of current lagging the voltage by 90°, A
When a capacitor is installed at the receiving end of the line, as shown in (Fig. 7.7 ) the resultant
voltage drop can be calculated approximately as
Figure ( 7.7) Voltage-phasor diagram for a feeder circuit of lagging power factor :(a) , (c)
with out (b) and (d) with shunt capacitor.
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[Distribution System for Residential area]Chapter 7
[]
where = reactive (or out-or-phase) component of current leading the voltage by 90°, A .
The difference between the voltage drops calculated by using Eqs. (7.5) and (7.6) is the voltage rise
due to the installa1ion of the capacitor and can be expressed as.
7.5.4 A Mathematical Procedure to Determine the Optimum Capacitor Allocation
The optimum application of shunt capacitors on distribution feeders to produce losses has bee
(Figure 7.8) shows a realistic representation of a feeder which contains a number of line segments
with a combination of concentrated (or lumped-sum) and uniformly distributed loads. Each line
segment represents a part of the feeder between sectionalizing devices, voltage regulators, or other
points of significance. For the sake of convenience, the load or line current and the resulting
loss can be assumed to have two components, namely,
(1) those due to the in-phase or active component of current and
(2) those due to the out-of-phase or reactive component of current
Since losses due to the in-phase or active component of line current are not significantly affected by
the application of shunt capacitors, they are not considered. This can be verified as follows.Assume that the losses are -caused by a lagging line current flowing
through the circuit resistance . Therefore it can be shown that
After adding a shunt capacitor with current , the resultants are a new line current.
, and a new power loss . Hence
Therefore the loss reduction as a result of the capacitor addition can be found
as
or by substituting Eq. (7.6) and (7.7) into Eq. (7.8).
Figure (7.8) Primary feeder with lumped-sum (or concentrated) and uniformlydistributed loads . And reactive current profile before adding capacitor.
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[Distribution System for Residential area]Chapter 7
[]
Thus only the out-of-phase or reactive component of line current , that is, I sin ,
should be taken into account for I'R loss reduction as a result of a capacitor
addition.
Assume that the length of a feeder segment is 1.0 pu length, as shown in
Fig. 7.7. The current profile of the line current at any given point on the feeder is
a function of the distance of hat point from the beginning end of the feeder.Therefore the differential loss of a dx differential segment located at a distance
x can be expressed as
Therefore the total R loss of the feeder can be found as
∫
∫
where
: is the total R loss of feeder before adding capacitor
: is the reactive current at beginning of feeder segment
: is the reactive current at end of feeder segmentR : is the total resistance of feeder segment
x : is the per unit distance from beginning of feeder segment
7.6 Loss Reduction Due to Capacitor Allocation
One capacitor bank The insertion of one capacitor bank on the primary , feeder causes a break in thecontinuity of the reactive load profile, modifies the reactive current profile , and consequently
reduces the loss, as shown in Figure (7.9).
Loss reduction with one capacitorFigure (7.9)
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[Distribution System for Residential area]Chapter 7
[6]
Therefore the loss equation after adding one capacitor bank can be found as before,
∫
∫
OR
[ ]
Thus the per unit power loss reduction as a result of adding one capacitor bank can be found from
or substituting Eq. and into Eq.
[ ]
or rearranging Eq. by dividing its numerator and denominator by so that
⁄ ⁄ ⁄ ⁄ ⁄ ⁄
If c is defined a: the ratio of the capacitive kilo volt amperes (c kVA) of the capacitor
bank to the total reactive load, that is,
then
and if is defined as the ratio of the reactive current at the end of the line segment to the reactive
current at the beginning of the line segment. that is,
Then
Therefore, substituting Eq. and into Eq. , the per unit power loss reduction can
be found as
show figure Loss reduction as a function of the capacitor-bank location and capacitor compensation ratio
for a line segment with uniformly distributed loads ( )
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[Distribution System for Residential area]Chapter 7
[5]
Figure (7.10) Loss reduction as a function of the capacitor-bank location and capacitorcompensation ratio for a line segment with uniformly distributed loads ( )
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[Distribution System for Residential area]Chapter 7
[]
7.6.1 Calculation Reduction of Power losses after add capacitor for Residential area
Active Power and Reactive Power for a Residential area
0
20000
40000
60000
1 2 3 4 5 6 7 8 9 10 11 12
P o w e r ( k w )
Month
Power Consume Per month
0
20000
40000
60000
80000
1 2 3 4 5 6 7 8 9 10 11 12
R e a c t i v e P o w e r ( k V A R )
Month
Reactive Power Per month
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[Distribution System for Residential area]Chapter 7
[]
;
√
√
∫
∫
∫
Where
x : per unit distance from beginning of feeder segment
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[Distribution System for Residential area]Chapter 7
[]
When c = 0.1 pu
; ;
;
;
When c = 0.2 pu
;
;
;
;
;
When c = 0.4 pu
;
;
;
;
;
When c = 0.6 pu
;
;
;
;
;
When c = 0.8 pu
;
; ;
;
;
When c = 1 pu
;
;
;
;
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[Distribution System for Residential area]Chapter 7
[7]
;
- CURVE FOR THE RESULTS
Figure (7.10) Loss reduction as a function of the capacitor-bank location and capacitor
compensation ratio for a line segment with uniformly distributed loads ( )
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[Distribution System for Residential area]Chapter 7
[]
7.6.1 Calculate the voltage drop per phase
√
*convert the , from to
So
743.33
2789
1234 I
559 -41.4
743.33
743.33
A
A
A
A
A
III
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[Distribution System for Residential area]Chapter 7
[]
voltage drop inside range stability not there any problem in this case
when we take location of capacitor equal 0.6 pu
-at c=0.1 given loss reduction by 0.21 pu
-at c=0.2 given loss reduction by 0.38 pu
-at c=0.4 given loss reduction by 0.65 pu
-at c=1 given loss reduction by 0.75 pu
-at c=0.6 given loss reduction by 0.8 pu
-at c=0.8 given loss reduction by 0.84 pu
choose capacitor depends on their cost and how much given power more than their cost
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[Conclusions]Chapter 8
38
Chapter 8
Conclusions
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[Conclusions]Chapter 8
3
Conclusions
At the end of this project we found the best type of conductors is Stranded Aluminum Conductor
with Stranded Steel Core (ACSR). The transmission line (R,L and C) are depend on configuration of
transmission line . we used bundle conductors to avoid the phenomena of corona and improve efficiency
by decreased the inductance (L) .
We study mechanical design of transmission line, we deal with main mechanical components of
overhead line and we study sag and tension between line between Alkudmi and Medical city.
We use software to calculate information of buses and power flow through lines of JAZAN
electrical equivalent network.
We use software and Z bus to calculate both types of short circuit occurs in buses and lines of
JAZAN electrical equivalent network. Short circuit analysis is performed to deduce the rating of
circuit breaker.
We reduce the system to an equivalent single machine infinite bus.
We check stability of system under study and we use PSS to enhance the stability .
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[Appendix]
58
1-Calculations of Transmission lines Parameters
Solution on Matlab program :...calculate mudium T.L
... where Vr=380
... type conductor ostrich
.... R=194.5 mohm/km XL=0.374 ohm/km
... Xc=0.227 Mohm/km GMR=7.01
.... Ir=490
... Vs=AVr+BIr % Is=CVr+DIr
... A=[1+(ZY/2) B=Z
... C=Y[1+(ZY/4)] D=[1+(ZY/2)
... F=60 L=100 KM =....................................................................... ....... enter value of Vr & Ir & R & XL & Xc o=0; L=80 ;Vr=(132*1000)/sqrt(3); R=7.376; XL=21.7; Xc=0.5848/1000; Z=R+XL*i; Y=Xc*i; for m=[.25:.25:1.50]Ir=(702-339.9941176i)*m; .................................................................. A=(1+(Z*Y/2));B=Z;C=Y*(1+(Z*Y/4));D=(1+(Z*Y/2));................................................................. Vs=(A*Vr)+(B*Ir);Is=(C*Vr)+(D*Ir);
MIr=abs(Ir) AIr=angle(Ir); AnIr=toDegrees('radians',AIr) PFr=cosd(AnIr) % angle of Ir VrL=Vr*sqrt(3); ................................................................. MVs=abs(Vs) VsL=MVs*sqrt(3) % magnitude of Vs AVs=angle(Vs);ANVs= toDegrees('radians',AVs) %angle of Vs ................................................................. MIs=abs(Is) % magnitude of Is AIs=angle(Is);
ANIs= toDegrees('radians',AIs) %angle of Is ................................................................. andegrees=(ANVs-ANIs); anradian= toradians('Degrees',andegrees); PFs=cos(anradian) ................................................................ Pr=sqrt(3)*MIr*VrL*PFr % power of PrPs=sqrt(3)*MIs*VsL*PFs % power of Ps.................................................. effeiciency=(Pr/Ps)*100 .......................................................... MA=abs(A); Vr0=VsL/MA;
regulation=((Vr0-VrL)/VrL)*100 end
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[Appendix]
5
2- Calculate Load Flow Using Matlab
% n v d Pg Qg PL QL type bus=[1 1.05 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1; 2 1.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 3; 3 1.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 3; 4 1.00 0.00 0.00 0.00 4.00 1.81 0.00 0.00 3; 5 1.02 0.00 15.0 0.00 7.79 0.00 0.00 0.00 2; 6 1.00 0.00 0.00 0.00 1.00 0.33 0.00 0.00 3; 7 1.00 0.00 0.00 0.00 2.34 0.77 0.00 0.00 3; 8 1.00 0.00 0.00 0.00 2.04 -0.33 0.00 0.00 3; 9 1.02 0.00 10.0 0.00 5.04 1.66 0.00 0.00 2; 10 1.00 0.00 0.00 0.00 4.34 1.43 0.00 0.00 3; 11 1.02 0.00 19.0 0.00 13.50 4.44 0.00 0.00 2; 12 1.00 0.00 0.00 0.00 2.902 1.405 0.00 0.00 3; 13 1.00 0.00 0.00 0.00 1.605 -0.72 0.00 0.00 3]; %From To R X B Tap line=[1 2 0.00 0.05 0 1. 0.;
2 3 0.00 0.05 0 1. 0.; 3 4 0.000815 0.0141 1.196 1. 0.; 4 6 0.00 0.05 0 1. 0.; 4 11 0.00375 0.0225 2.64 1. 0.; 5 6 0.00402 0.03374 0.04 1. 0.; 6 7 0.005 0.0214 0.0174 1. 0.; 6 8 0.029 0.017 0.0154 1. 0.; 6 10 0.0174 0.007225 0.006 1. 0.; 7 9 0.00415 0.0267 0.00438 1. 0.; 4 12 0.00741 0.006891 1.724 1. 0.; 6 13 0.01058 0.02585 0.4076 1. 0.;]; [bus_sol,Line_flow]=loadflow(bus,line,0.001,30,0.5,1.5,1, 'y',1); vproject=bus_sol(:,2) % voltage
pproject=bus_sol(:,4) %power eproject=bus_sol(:,3)*pi/180
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[Appendix]
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3- Calculations electric torque ,field voltage equation and terminal voltage magnitude of get stability
network using matlab and Simulink
M=8.2; Tdo=4.91;
D=0; Xd=0.5975; xd=5.875;% note x dash Xq=5.625; %---------------------------------------- Z=-0.00273+0.0747i; Y=16.913-5.074i; R=-0.00273; X=0.0747; G=16.913; B=-5.074; Vo=0.9583; ANo=(-6.88*pi)/180; Vt=1.02; ANt=((-0.837*pi)/180); Pe=19; Qe=3.88; %--------------------------------------------------- A=1+(Z*Y); C1=real(A) C2=imag(A) R1=R-(C2*xd) X1=X+(C1*Xq) R2=R-(C2*Xq) X2=X+(C1*xd) Ze=(R1*R2)+(X1*X2) Yd=((C1*X1)-(C2*R2))*Ze Yq=((C1*R1)+(C2*X2))*Ze
%---------------------------------------------------- Vd=(Pe*Vt)/(Pe^2+(Qe+(Vt^2/Xq))^2)^0.5 Vq=(Vt^2-Vd^2)^0.5 Iq=Vd/Xq Id=(Pe-(Iq*Vq))/Vd Eq=Vq+(xd*Id) %---------------------------------------------------- Fdq=(Vo/Ze)*[-R2 X1;X2 R1]*[cos(ANo);sin(ANo)]; Fd=Fdq(1:1) Fq=Fdq(2:2) %----------------------------------------------------- K12=[0;Iq]+[Fd Fq;Yd Yq]*[(Xq-xd)*Iq;Eq+((Xq-xd)*Id)]; K1=K12(1:1) K1=K12(2:2) %------------------------------------------------------- K3=1/[1+((Xd-xd)*Yd)] K4=(Xd-xd)*Fd %------------------------------------------------------- K56=[0;(Vq/Vt)]+[Fd Fq;Yd Yq]*[-xd*Vq/Vt;Xq*Vd/Vt]; K5=K56(1:1) K6=K56(2:2) %----------------------------------------------------- Vind=(C1*Vd)-(C2*Vq)-(R*Id)+(X*Iq) Vinq=(C2*Vd)+(C1*Vq)-(R*Iq)-(X*Id)
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References
References
1) J. Duncan Glover, Mulukutla S. Sarma, Thomas J. Overbye, “Power System Analysis and Design”,
Global Engineering, 2012, Fifth Edition.
2) Das, “Electrical Power Systems”, 2006, New Age International (P) Ltd., Publishers
3) M.E. El-Hawary, “Introductionto Electrical Power Systems”, A John Wiley & Sons, Inc., Publication,
2008.