design of transmission line between alkudmi and medical city of jazan

7/25/2019 Design of Transmission Line Between ALKUDMI and Medical City of Jazan

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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



[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




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.




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




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




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




7

Chapter 1

Introduction




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.




9

The Figure (1.1) shows the equivalent electrical network of Jazan .

Figure (1.1) : equivalent electrical network of Jazan



Electrical Performance of Transmission LineChapter 2

[01]

Chapter 2

Electrical Performance of

Transmission Lines




[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




[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.




[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.




[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




[0]

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.




[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 .




[0]

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).




[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 .




[0]

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.




[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)




[0]

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.




[]

(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)




[]

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)




[]

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




[]

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 :

, ,

,

,

√ , √

√ √




[]

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

|

√ |

|




[]

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)




[]

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.




[]

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



[Mechanical Design of Transmission Line]Chapter 3

03

Chapter 3

Mechanical Design of

Transmission Line




0

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.




0

(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.




00

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.




0

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

⁄




0

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




0

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.




0

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




0

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).




0

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




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




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)




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)




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




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

( )



[Power Flow Analysis]Chapter 4

4

Chapter 4

POWER FLOW

ANALYSIS




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.




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.




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 ) :




)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)




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




4

4.7 Load flow by using Power world Simulator Program :

Figure (4.1) equivalent electrical network of Jazan



[Fault System Analysis]Chapter 5

25

Chapter 5

FAULT SYSTEMANALYSIS




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 .




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




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)




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.




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)




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)




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 :




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



[Power System Stability]Chapter 6

16

Chapter 6

Power System Stability




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




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.

-

+

+

(

) (

)




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.



[Distribution System for Residential area]Chapter 7

[56]

Chapter 7

Distribution System

for Residential area




[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




[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.




[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).




[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




[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.




[]

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.




[]

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.




[]

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.




[]

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)




[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 ( )




[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 ( )




[]

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




[]

;

√

√

∫

∫

∫

Where

x : per unit distance from beginning of feeder segment




[]

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

;

;

;

;




[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 ( )




[]

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




[]

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=1 given loss reduction by 0.75 pu



choose capacitor depends on their cost and how much given power more than their cost



[Conclusions]Chapter 8

38

Chapter 8

Conclusions



[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 .



[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



[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



[Appendix]

5

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)



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.

design of transmission line between alkudmi and medical city of jazan

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