2. 2
AN-Najah National University
Faculty of Engineering
Electrical Engineering Department
Introduction to Graduation Project
Optimum Performance of Tulkarim Governorate network
& Sarra Connection Point
ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ
Supervisor:
Dr. Imad Ibrik
Prepared by:
Woroud turabi
Ahmad nasralla
3. 3
Acknowledgment:
Our first and greatest thanks are to Allah. He above all was, is,
and will be the source of help and guidance that counts. His help kept us
going through many frustrations and His guidance brought us back on track
when our frustrations tended to drive us astray.
Next, we would like to express our deep gratitude to our
supervisor Dr. Imad Ibrik for his effort and encouragement throughout the
preparation of this project.His Knowledge and endless support was a great
asset from which we learned plenty. He was very enthusiastic about the
project, which gave us a motive to work harder and harder.
We would like to thank friends who did their best whenever we
needed or asked for anything. No matter what the task was or how busy
they were, they always found the time.
Also we would like to thank our families that help us very much and
they were very generous with us.
It is difficult to acknowledge everyone who was involved in
preparation of this project by name. Nevertheless we appreciate their help
no matter how simple it might have been.
Finally, we owe our colleagues great thanks for supporting us
through our college years and for making those years the best years of our
life.
4. 4
DISCLAIMER
This report was written by students at the Electrical and
Telecommunication Engineering Department, Faculty of
Engineering, An-Najah National University. It has not been
altered or corrected, other than editorial corrections, as a result of
assessment and it may contain language as well as content
errors. The views expressed in it together with any outcomes and
recommendations are solely those of the students. An-Najah
National University accepts no responsibility or liability for the
consequences of this report being used for a purpose other than
the purpose for which it was commissioned.
5. 5
Contents:
List of tables and figures …………………………………………………………
IEC standards………………………………………………………………………………………..
Nomenclature or list of symbols………………………………………………
Abstract…………………………………………………………………………………..
Introduction ……………………………………………………………………………
Chapter 1: Tulkarim networks & Sarra connection point
Description of the networks………………………………………………………
Sourceof information……………………………………………………………….
Chapter 2: Transmission lines…
Electrical power transmission………………………………………
Type of transmission line…………………………………………….
Chapter 3: analysis for the existing networks
One line diagramfor networks…………………………………………………
Analysis of the networks……………………………………………………………
Problems in the networks…………………………………………………………
Chapter4: analysis of supply sarra connection point and
tulkarim from central substation
Improving thenetworkes …………………………………………………………………………
1. Tap changing
6. 6
2. Power factor improvement
3. Changing over load transformer
Chapter5: Mechanical design of the network
Main Components of overhead lines…………………………………………………….
1. Conductors
2. Supports
3. Insulators
ImportantPoint………………………………………………………………………………………
Sag in Overhead Lines……………………………………………………………………………..
Calculation of Sag………………………………………………………………………………….
References………………………………………………………………………………...
Appendix…………………………………………………………………………………..
List of tables:
Table 1.1A: transformersof University station………………………………………………………….
1.2A: transformersof al karakonstation…….…………………………………………………..
1.1B: transformersof tulkarim1……………………………………………….…………………..
1.2B: transformersof tulkarim2…………………………………………..……………………….
1.3: R & X of transformers…………………………………………………………………………...
1.4A: the loadandpowerfactor of each transformer…………………………………………
1.4B: the loadand powerfactorof each transformer…………………………………………
Table 2.1: R&X of the ACSR ………………………………………………………………
2.2: R&X of the XLPE Cu……………………………………………………………….
7. 7
2.3: R&X of the XLPE Al…………………………………………………………………..
Table 3.1A: rated powerof eachtransformer…………………………………………………..
3.1B: ratedpowerof each transformer…………………………………………………..
3.2A: full loaddescription……………………………………………………………………
3.2B: full loaddescription……………………………………………………………………
3.3A: summaryof total generation,loading&demand……………………
3.3B: summaryof total generation, loading&demand……………………
Table 4.1A : summaryof total generation,loading&demand……………………
4.1B: summaryof total generation,loading&demand……………………
4.2A(TAPCHANGE) : summaryof total generation,loading&demand…………….
4.2B(TAP CHANGE) : summaryof total generation,loading&demand…………….
4.3: The penaltiesof powerfactor………………………………………………………………………..
4.4A(POWERFACTOR) : summaryof total generation,loading&demand…………
4.4B(POWER FACTOR) : summaryof total generation,loading& demand…………
4.5 showsthe transformerswhichare neededtobe bought……………………………………
4.6 summarizesthe analysisresultsafterchangingtransformers………………………………….
Table 5.1: Numberof steel towersandtrusswhichwe need…………………………………………………
5.2: Numberof insulatorsthatwe needineachtype……………………………………………………
8. 8
List of figures:
Figure a. a: the growthpatternin WestBank,Gaza Stripand the total Palestine forecast………….
3.1A: distributionof transformersanddistance betweenthem……………
3.1B: distributionof transformersanddistance betweenthem…………………………….
3.2A: one line diagramforSarra connectionpoint’stransformer………..
3.2B: one line diagramforTulkarimnetworks…………………………………
3.3A: firstrun for the network shows(s,v & pf)………………………………
3.3B: firstrun forthe networkshows(s,v & pf)………………………………
4.1 : 5% tap changer………………………………………………………………………………………………………..
5.1:steel cordaluminumconductor……………………………………………………………………………………..
5.2: pintype insulators……………………………………………………………………………………………………….
5.3: suspensioninsulators…………………………………………………………………………………………………..
5.4: straininsulator……………………………………………………………………………………………………………..
5.5: trusswitharm in front…………………………………………………………………………………………………….
5.6: tower……………………………………………………………………………………………………………………………..
5.7: sag inoverheadlines……………………………………………………………………………………………………….
5.8: sag whensupportsare at equal levels…………………………………………………………………………….
5.9: sag whensupportsare at unequal levels………………………………………………………………………….
5.10: Effectof windand ice loading……………………………………………………………………………………..
10. 10
IEC standards
This is an incomplete list of standards published by the
International Electrotechnical Commission(IEC) 1]:
IEC 60038 IEC StandardVoltages
IEC 60044 Instrumenttransformers
IEC 60028 International standardof resistance forcopper
IEC 60076 Powertransformers
IEC 60085 Electrical insulation
IEC60228Conductorsofinsulated[2]
IEC209 Aluminumstrandedconductorsteelreinforced [3]
IEC 60871 capacitorbank [4]
11. 11
Nomenclature or list of symbols:
A ampere(s)
V volt(s)
W watt(s)
KA kiloampere(s)
KV kilovolt(s)
Km kilometer(s)
MVA megavolt ampere(s)
MVAR megavolt-ampere(s) reactive
MW megawatt
ACCR aluminum conductor composite reinforced
Xlpe cross-linked polyethylene power cable
m meter
mm2 millimeter square
mm millimeter
12. 12
{Abstract}:
The important aspects to be covered in this project are preparing the initial data
for Tulkarim Governoratenetwork & Sarra Connection Point and subjectto a load
flow study using modern softwarelike ETAP to improvethe voltage level and the
power factor and reduce the electrical losses by Reconnection the two networks
with Sarra electricity converter station which converts from33Kv to 11Kv instead
of (336.6) KV.
The objectives of the project are:
To be familiar with TulkarimGovernoratenetwork & Sarra
Connection Point
To improvethe voltage level and decreasethe real power losses
To increase the reliability of the networks
To connect the two networks( Tulkarimand Sarra connection point)
to Sarra electricity converter station (16133 KV) directly without
relying on Israelinational electricity company
Giving recommendation for the best systemto be used in Sarra
connection point and in Tulkarim (e.g reconnection, change
transformers).
Giving recommendation for the design to be used in Sarra connection
point and in Tulkarim
In order to do these objectives these method will be followed :
Built the one line diagramfor ETAP program
Collect the data for the networks including all parameters
Load flow analysis and study for networks by Detect problems in
networks
13. 13
Energy sector in Palestine
Energy sector in Palestine faced many difficulties becauseof occupation. Till now
there is no unified power systemin Palestine. Most of electrical energy depends
on IEC Company except Jericho which connected with Jordan and Gaza to Egypt
(17MW) through the interconnection project. The only generation plant is in Gaza
with generating capacity of 140MW. Distribution companies take the role of
distributing electricity in the different regions of Palestine.
The average annual growth rate of energy demand in westbank is 6.4%, and in
Gaza is 10% from1999 to 2005. [5]
Figure a. a: the growth patterninWestBank, Gaza Stripand the total Palestineforecast
15. 15
Power system
The power system is complicated electrical networks used to supply, transmit, and use
electrical energy .the networks supply s towns containing houses hospitals industrial
region called CRID. the grid is contains generators that supply the power ,the
transmission system that carries the power from the generating center to the load center
and the distribution system that feeds the power to the nearby home and industries . the
majority of these system rely upon three-phase AC power –the standard for large-
scale power transmission and distribution across the modern world .specialized power
systems that do not always rely three-phase AC power are found in aircraft ,electric rail
systems , ocean linear and automobiles 1]
Power flow study:
In power engineering, the power flow study (also known as load-flow study) is an
important tool involving numerical analysis applied to a power system. Unlike traditional
circuit analysis, a power flow study usually uses simplified notation such as a one-line
diagram and per-unit system, and focuses on various forms of AC power (ie: reactive,
real, and apparent) rather than voltage and current. It analyzes the power systems in
normal steady-state operation. There exist a number of software implementations of
power flow studies. In addition to a power flow study itself, sometimes called the base
case, many software implementations perform other types of analysis, such as short-
circuit fault analysis and economic analysis. In particular, some programs use linear
programming to find the optimal power flow, the conditions which give the lowest cost
per kilowatt generated.
The great importance of power flow or load-flow studies is in the planning the future
expansion of power systems as well as in determining the best operation of existing
systems. The principal information obtained from the power flow study is the magnitude
and phase angle of the voltage at each bus and the real and reactive power flowing in
each line. Commercial power systems are usually too large to allow for hand solution of
the power flow. the first step in the load flow study is simulate the power system by the
one line diagram and collect the data and specification of all power system contents and
then translate this system to per unit circuit in order simplifies it ,this happened by
choose abases value in the power system .[6]
16. 16
Bus classification:
Load bus: In these buses no generators are connected and hence the generated
real power PGi and reactive power QGi are taken as zero. The load drawn by these
buses are defined by real power -PLi and reactive power -QLi in which the negative
sign accommodates for the power flowing out of the bus. This is why these buses
are sometimes referred to as P-Q bus. The objective of the load flow is to find the
bus voltage magnitude |Vi| and its angle δi.[7]
Generator bus or voltage controlledbus:Theseare the buses wheregenerators
are connected. Thereforethe power generation in such buses is controlled
through a prime mover while the terminal voltage is controlled through the
generator excitation. Keeping the input power constant through turbine-governor
control and keeping the bus voltage constant using automatic voltage regulator,
we can specify constant PGi and | Vi | for these buses. This is why such buses are
also referred to as P-V buses. Itis to be noted that the reactive power supplied by
the generator QGi depends on the systemconfiguration and cannot be specified in
advance. Furthermorewe haveto find the unknown angle δi of the bus voltage.
[7]
Slack (swing) bus:Usually this bus is numbered 1 for the load flow studies. This
bus sets the angular reference for all the other buses. Sinceit is the angle
difference between two voltage sources that dictates the real and reactive power
flow between them, the particular angle of the slack bus is not important.
However it sets the reference against which angles of all the other bus voltages
are measured. For this reason the angle of this bus is usually chosen as 0°.
Furthermoreit is assumed that the magnitude of the voltage of this bus is
known.[7]
19. 19
tthheessee rroooottss ooff aa ssiimmppllee ffuunnccttiioonn ssuucchh aass:: ff((xx)) == xx22
--44 ssiimmppllyy bbyy sseettttiinngg tthhee ffuunnccttiioonn
ttoo zzeerroo,, aanndd ssoollvviinngg::
FF((xx)) == xx22
--44 == 00
((xx++22))((xx--22)) == 00
xx == 22 oorr xx == --22
The Newton-Rap son method uses an iterative process to approach one root of a
function. The specific root that the process locates depends on the initial,
arbitrarily chosen x-value.
Here, xn is the currentknown x-value, f(xn) represents the value of the function at
xn, and f'(xn) is the derivative (slope) at xn. xn+1 representthe next x-value that you
are trying to find. Essentially, f'(x), the derivative represents f(x)/dx(dx = delta-x).
Therefore, the term f(x)/f'(x) represents a value of dx.
TThhee mmoorree iitteerraattiioonn tthhaatt aarree rruunn,, tthhee cclloosseerr ddxx wwiillll bbee ttoo zzeerroo ((00))..
The Newton-Raphson method does not always work, however. Itruns into
problems in several places. First, consider the above example. Whatwould
happen if we chosean initial x-valueof x=0? We would have a "division by zero"
error, and would not be able to proceed. You may also consider operating the
process on the function f(x) = x1/3
, using an initial x-valueof x=1. Do the x-values
converge? Does the delta-x decrease toward zero (0)? 1
So, how does this relate to chemistry? Consider the van der Waals equation found
in the Gas Laws section of this text. Assuming that we have a set number of moles
of a set gas, not under ideal conditions, we can use the Newton-Raphson method
to solve for one of the three variables (temperature, pressure, or volume), based
on the other two. To do this, we need to use the van der Waals equation, and the
derivative of this equation, both seen below [6].
20. 20
As you can see, the Van der Waals equation is quite complex. It is not possibleto
solveit algebraically, so a numerical method must be used. The Newton-Raphson
Method is the easiest and most dependable way to solveequations like this, even
though the equation and its derivative seem quite intimidating. Depending on the
conditions under which you are attempting to solvethis equation, severalof the
variables may be changing. So, it may be necessary to use partial derivatives. For
the purposes of this example, we are assuming thatpressure, temperature, and
volume are the only things changing, and that these values are all functions of
time. This avoids the useof a partial derivative; we simply differentiate all
variables with respect to time, as shown above. Somealgebraic manipulation of
the equation and/or its derivative may be needed depending on the specific
problem to be solved. Itis assumed that all of the variables but one are specified;
that variable is used in the expression for "xn+1" that Newton's method uses.
Performing Newton's method on this equation successfully would givea value of
that variable which gives a solution when the other variables are held constant at
the values you specified.
Real Power Losses
The real power losses: is the loss of the systembecauseof the voltage drop on the
busses.
In the load buses we must feed the load the power which is needed, put if the
voltage on the buses is very low the current become higher than beforeas the
following:
P = V*I *P.f
P is constant, then:
If the voltage (V) is decreased the current (I) is increased,
∆P = 3*I2
* R
22. 22
Components of power system
1.Power station
The power station of a power system consists of a prime mover, such as a
turbine driven by water, steam, or combustion gases that operate a system of
electric motors and generators. Most of the world's electric power is generated in
steam plants driven by coal, oil, nuclear energy, or gas. A smaller percentage of
the world’s electric power is generated by hydroelectric (waterpower), diesel, and
internal-combustion plants.[8]
2.Transformer
electric power systems transformer is a static device which transforms electrical
energy from one circuit to another without any direct electrical connection and
with help of mutual induction between to windings .it transforms power from one
circuit to another without changing its frequency but may be in different voltage
level .the transformer can work either as a step up or a step down voltage or
current. We have two types of transformers used in power system first one is
power transformer and the second is distribution transformer, power transformer
are used in the high voltage level usually in power station and in substation and
the distribution transformer are used in the medium and low voltage side
(consumer side) [8]
3.Transmission line
A transmission line is a material medium or structure that forms a path for
directing the transmission of energy from one place to another, such as
electromagnetic waves or acoustic waves as well as electric power transmission.
It use to carry the power from power station to the load that represent the
consumer ,it divided to long line and medium line and short line .
4.Load
Which it’s both reactive and real power and both of them are specified and both
voltage magnitude and angle are determined by the computer as part of solution, which
they’re both voltage magnitude and real power are specified and we will determine the
reactive power and angle by the computer program as part of solution so the generation
busses is one of the method to solve the parameter of the complex network. [8]
23. 23
Chapter 1
Tulkarim Governorate networks & Sarra Connection
Point ….
1.1 description of the networks
Sarra connection point is fed by Israel Company, the capacity of this connection
point is 20MVA (rated) and the rated voltage is 33KV. This connection point
contains 92 distribution transformers with differentrating power depending on
the region (or the residential) and the consumption of energy. [9]
Table1.1A: transformers of University station
Transformer number Rating Rated power (KVA)
Tr1 southern 3311-6.6 10000
Tr2 madakh al- jneed 116.6-0.4 400
Tr3 amena saaed 116.6-0.4 160
Tr4 khalele 116.6-0.4 400
Tr5 jalal yaseen 116.6-0.4 400
Tr6 tayba 1 116.6-0.4 400
Tr7 tayba 2 116.6-0.4 400
Tr8 al ameria 116.6-0.4 630
Tr9 eskan almohandesen 116.6-0.4 250
Tr10 eskan Shinar 116.6-0.4 250
Tr11 bet wazan 116.6-0.4 250
Tr12 hajez 116.6-0.4 400
Tr13 jneed 116.6-0.4 400
Tr14 orabee 116.6-0.4 400
Tr15 alferdaws 116.6-0.4 630
Tr16 khateeb 116.6-0.4 630
Tr17 afonneh 116.6-0.4 630
Tr18 seha 116.6-0.4 400
Tr19 jaber 116.6-0.4 400
Tr20 msjed al makhfeha 116.6-0.4 400
Tr21 kamal jnblat 116.6-0.4 630
26. 26
Tr50 tel gharbee 11-6.60.4 400
Tr51 tareeq sarra 11-6.60.4 400
Tr52 tel madakha 11-6.60.4 250
Tr53 raze 11-6.60.4 400
Tr54 dardok 11-6.60.4 400
Tr55 joharee 11-6.60.4 400
Tr56 shohadaa 11-6.60.4 400
Tr57 qteshat 11-6.60.4 250
Tr58 el qaser 11-6.60.4 1000
Tr59 shanaa 11-6.60.4 250
Tulkarim Governorateis fed by IsraelCompany, and its contain two networks
1&2 the capacity of the firstone is 13MVA (rated) and the rated voltage is 22KV,
the second is 11.968 MVA (rated) and the voltage is 22KV also . These networks
contain 101 distribution transformers with differentrating power depending on
the region (or the residential) and the consumption of energy. 1
Table1.1B: transformers of tulkarim1
Transformers
number
RatingRated power
(KVA)
63022/0.41خضوري
63022/0.4رقم 1بئر
63022/0.42خضوري
63022/0.4الناصر عبد جمال 1دوار
63022/0.4الناصر عبد جمال 2دوار
63022/0.4رقم 2بئر
40022/0.4خضوري دوار
40022/0.4الدلهوم
40022/0.4د.ثابت
25022/0.4الجنوبي الحي
40022/0.4هواش
30. 30
1.2 source of information
When we stabilize the idea we searched for sources of information about what
we intend to do:
At first wewent to the municipality of Tulkarim and Northern Electric Distribution
Company(NEDCO) , and we got the information about the transformer ( rated
power) and the transmission lines length (the cross sectionalarea and the type
),and we measured the load of each transformer practically by ourselves also we
measured the impedances of each transmission line .
To calculate the load for each transformer wecan usethese formulas:
𝑆 𝑙𝑜𝑎𝑑 = 𝑙𝑜𝑎𝑑 𝑓𝑎𝑐𝑡𝑜𝑟 ∗ 𝑆 𝑡𝑟
𝑃 = 𝑆 ∗ 𝑐𝑜𝑠𝜃
𝑄 = √ 𝑆2 − 𝑃2
The average demand load factor in our network is 50% that means the average
load to the maximum load ratio is 50% which consider as a very good operating
load factor.
And the table below shows thevalue of the load for each transformer and the
power factor of the load.
Table1.4A: the load and power factor of each transformer
Transformer number Rated power (KVA) p(Kw) Q(Kvar) Power factor
Tr1 southern 5000 - - -
Tr2madakh al- jneed 200 180 87.18 90
Tr3 amena saaed 80 72 34.87 90
Tr4 khalele 200 180 87.18 90
Tr5 jalal yaseen 200 180 87.18 90
Tr6 tayba 1 200 180 87.18 90
37. 37
Chapter 2:
Transmission lines…
1.1Electric-power transmission is the bulk transfer of electrical energy, from
generating power plants to electrical substations located near demand centers. This is distinct
from the local wiring between high-voltage substations and customers, which is typically
referred to as electric power distribution. Transmission lines, when interconnected with each
other, become transmission networks. The combined transmission and distribution network is
known as the "power grid" .
The system:
Most transmission lines are high-voltage three-phase alternating current (AC), although single
phase AC is sometimes used in railway electrification systems. High-voltage direct-current
(HVDC) technology is used for greater efficiency at very long distances (typically hundreds of
miles (kilometers)), or in submarine power cables (typically longer than 30 miles (50 km)).
HVDC links are also used to stabilize and control problems in large power distribution networks
where sudden new loads or blackouts in one part of a network can otherwise result in
synchronization problems and cascading failures.[11]
Diagram of an electric power system
38. 38
1.2types of transmission lines that used in
this network
1. Overhead transmission..
High-voltage overhead conductors are not covered by insulation. The conductor material is
nearly always an aluminum alloy, made into several strands and possibly reinforced with steel
strands. Copper was sometimes used for overhead transmission, but aluminum is lighter, yields
only marginally reduced performance and costs much less. Overhead conductors are a
commodity supplied by several companies worldwide. Improved conductor material and shapes
are regularly used to allow increased capacity and modernize transmission circuits. Conductor
sizes range from 12 mm2
to 750 mm2
(1,590,000 circular mils area), with varying resistance and
current-carrying capacity. Thicker wires would lead to a relatively small increase in capacity due
to the skin effect, that causes most of the current to flow close to the surface of the wire.
Because of this current limitation, multiple parallel cables (called bundle conductors) are used
when higher capacity is needed. Bundle conductors are also used at high voltages to reduce
energy loss caused by corona discharge .[11]
In these networks, the type of overhead transmission lines that used are ACSR (aluminum
conductor steel rain forced) ,the resistance and reactance of this conductor shows in the below
table
ACSR cable R(ohmsKm) X(ohmsKm)
120 mm2 0.219 0.269
95mm2 0.301 0.322
50mm2 0.543 0.333
Table 2.1: R&X of the ACSR
39. 39
2. Underground transmission
Electric power can also be transmitted by underground power cables instead of overhead power
lines. Underground cables take up less right-of-way than overhead lines, have lower visibility,
and are less affected by bad weather. However, costs of insulated cable and excavation are
much higher than overhead construction. Faults in buried transmission lines take longer to
locate and repair. Underground lines are strictly limited by their thermal capacity, which permits
fewer overloads or re-rating than overhead lines. Long underground AC cables have significant
capacitance, which may reduce their ability to provide useful power to loads. [11]
In these networks, the type of overhead transmission lines that used are XLPE Cu ,
XLPE Al ,the resistance and reactance of this conductor shows in the below table
XLPE Cu R(ohmsKm) X (ohmsKm)
240mm2 0.754 0.109
120mm2 0.196 0.117
95mm2 0.41 0.121
50mm2 0.387 0.138
Table 2.2: R&X of XLPE Cu
XLPE Al R(ohmsKm) X(ohmsKm)
95mm2 0.32 0.542
Table 2.3: R&X of XLPE Al
We measured the length of each transmission line
40. 40
Chapter 3
Analysis for the existing networks…..
3.1 One line diagram for the networks
In this chapter we will study the networks beforethe improvement and
connecting the district,to find the losses of the power in the networks and to see
if the networks haveany problems like (over load on any transformer ,and the
drop voltage and power factor and voltage level or any problem that happened in
these networks )
Fromthe data in chapter one and two weget the information about our networks
to make the one line diagramfor the networks, thetables below will show the
required data for the one line diagram.
1. Number of transformers
Table 3.1A: rated power of each transformer
Transformer number Rated power (KVA)
Tr1 southern 10000
Tr2madakh al- jneed 400
Tr3 amena saaed 160
Tr4 khalele 400
Tr5 jalal yaseen 400
Tr6 tayba 1 400
Tr7 tayba 2 400
Tr8 al ameria 630
Tr9 eskan almohandesen 250
Tr10 eskan Shinar 250
Tr11 bet wazan 250
Tr12 hajez 400
Tr13 jneed 400
Tr14 orabee 400
Tr15 alferdaws 630
Tr16 khateeb 630
Tr17 afonneh 630
41. 41
Tr18 seha 400
Tr19 jaber 400
Tr20 msjed al makhfeha 400
Tr21 kamal jnblat 630
Tr22 etsalat 1 400
Tr23 etsalat 2 630
Tr24 reyada 630
Tr25 saydleh 630
Tr26 funon 1500
Tr27 tamred 1500
Tr28 oloom 1500
Tr29 hndesah 1500
Tr30 sejen jneed 630
Tr31 seefe 400
Tr32 eskan shinar khalf jneed 250
Tr33 al bydar 400
Transformers number Rated power (KVA)
Tr1 karakon1 10000
Tr2 karakon2 10000
Tr3 karakon feeder 630
Tr4 heteen 400
Tr5 yasmeen hotel 630
Tr6 bab saha 630
Tr7 malhes 630
Tr8 aqaree bank 630
Tr9 Jordan bank 630
Tr10 othmanee 250
Tr11 revolee 630
Tr12 Palestine bank 400
Tr13 abo salha 630
Tr14 alkonee 630
Tr15 alsook alekhdar 500
Tr16 alhewaree 630
Tr17 tokan 630
Tr18 shakaa 400
Tr19 malhees 400
Tr20 alenjeehe 500
Tr21 kalbone1 630
Tr22 ksheka 400
Tr23 krom ashoor 250
Tr24 aeen al asel 400
42. 42
Tr25 madakhet ras eleen 630
Tr26 salah deen 400
Tr27 Samsung 630
Tr28 el basha 630
Tr29 omer ben el aas 400
Tr30 kalboneh 2 250
Tr31 madakhet 24 630
Tr32 abo raed 630
Tr33 blaza 250
Tr34 takhasosee 1000
Tr35 sharea 24 630
Tr36 Kazan janobee 400
Tr37 jneed janobee 400
Tr38 karajat el jameaa 630
Tr39 maktabet el jameaa 400
Tr40 jameaa nor 400
Tr41 eskan naqabat 400
Tr42 sharea tel 630
Tr43 arade shinar 250
Tr44 Nablus jadeeda 630
Tr45 Iraq boreen 1 250
Tr46 Iraq boreen 2 250
Tr47 tel sharqee 400
Tr48 tel lehef 250
Tr49 tel el bald 250
Tr50 tel gharbee 400
Tr51 tareeq sarra 400
Tr52 tel madakha 250
Tr53 raze 400
Tr54 dardok 400
Tr55 joharee 400
Tr56 shohadaa 400
Tr57 qteshat 252
Tr58 el qaser 1222
Tr59 shanaa 252
46. 46
2. distance between transformers
By using the AutoCAD program for Sarra connection point and the single line
diagramfor Tulkarimnetwork, we specified and draw the location for each
transformer, also wedraw the distribution of the transmission lines and we show
the type of each line (overhead and underground line), after that we measured
the length of each transmission line as we see in the below picture:
Figure 3.1A: distribution of transformers and distance between them
50. 50
After we show the distribution of transformers on the single line diagramand the
AutoCAD and after collecting all information, we design the one line diagramfor
these networks as we see in the pictures below
Figure 3.2A: one line diagram for Sarra connection point’s transformer
58. 58
After we doing this analyzing for the network, weobtained the results about the
powers and losses which shown in the next summery table
59. 59
Table 3.3A summary of total generation, loading & demand
Fromthe abovetable we can see the value of real power for the swing bus
(source) is equal to 20.820 MW, this is the real power that consuming by this
network and the reactive power for the network is equal to 10.213MVAr.
We note that the apparentpower of the network which is 23.190MVA
Also we note that the losses in the network for the real power is equal to
0.443MW and for reactive power is equal to 0.366 MVAr.
63. 63
ثابت 2مستشفى 346 0.4 312 151 151 90 0.389 96.893
العدويه 173 0.4 156 75.498 90 0.39 97.286
دعباس مجمع 173 0.4 156 75.498 90 0.39 97.592
التاج مجمع 346 0.4 312 151 90 0.389 97.450
الخاروف مجمع 173 0.4 156 75.498 90 0.385 97.228
الشاهد 416 0.4 374 181 90 0.388 97.059
المحافظه 346 0.4 312 151 90 0.389 97.195
المقاطعه 416 0.4 374 181 90 0.388 97.129
The sourcerated power for the second network is equal to 10.376 MW and this
value of power is suitable and enough for this network. The pictures below shows
the firstrun for these networks atmax case in ETAP programthat shows the
distribution of power and its direction between buses and the voltage for each
bus and the power factor.
66. 66
After we doing this analyzing for the two networks, weobtained the results about
the powers and losses which shown in the next summery tables
Table 3.3B summary of total generation, loading & demand
Fromthe abovetable we can see the value of real power for the swing bus
(source) is equal to 15.926 MW which is more, this is the real power that
consuming by the firstnetwork and the reactive power for this network is equal
to 7.845MVAr.
We note that the apparentpower of the network which is 17.753MVA and this
value unacceptable.
Also we note that the losses in the network for the real power is equal to 0.88MW
and for reactive power is equal to 0.558 MVAr
67. 67
Fromthe abovetables we can see the value of real power for the swing bus
(source) is equal to 10.376 MW, this is the real power that consuming by the
second network and the reactive power for this network is equal to 5.061MVAr.
We note that the apparentpower of the network which is 11.554MVA and this
value does not makea problemfor consuming power.
Also we note that the losses in the network for the real power is equal to
0.233MW and for reactive power is equal to 0.149 MVAr
68. 68
3.3 problems in the net work
voltages
The voltage mustbe in the range:
0.95 V nominal <V < 1.05 V nominal
After we doing the analyzing for the networks on the ETAP programwe see
that the voltages in somebuses are not located desired range
power factor
The poor factor in many regions is low and we are looking to improveit to
92 and more
there are 3 transformers (T29,T38,T55) in Tulkarimnetwork 1 that
connectedtoover load
The capacity of T29 is 250 KVA and the load that connected to this
transformer has a power equal to 238 KVA.
The load that connected to the transformer #29 is 95.2% fromthe total
load and this value is abovethe rangeof the load that allotted for each
transformer, and the rangefor the load that mustconnect to the
transformer between (50%-75%).
69. 69
The capacity of T38 is 250 KVA and the load that connected to this
transformer has a power equal to 271 KVA.
The load that connected to the transformer #38 is 108% fromthe total load
and this value is above the rangeof the load that allotted for each
transformer
The capacity of T55 is 400 KVA and the load that connected to this
transformer has a power equal to 404 KVA.
The load that connected to the transformer #55 is 101% fromthe total load
and this value is above the rangeof the load that allotted for each
transformer
For more information you can see the appendix
70. 70
Chapter4:
Analysis of supply Sarra connection point and
Tulkarem from central substation
At this stage of our graduation project we will Study the new condition of the two
networks (Tulkarimand Sarra connection point) after connecting them to Sarra
electricity distribution substation (16133) KV directly without relying on Israeli
national electricity company. Then we will improvethe voltage level and decrease
the real power losses and increase the reliability of the networks.
After the analysis of Sarra connection point and tulkaremnetworks after we
connecting them to Sarra electricity distribution substation (16133) KV also many
problems in the network appears as we mentioned before
For more details you can see the appendix
and the following tables will show us the conditions of the networkes.
Table 4.1A : summary of total generation, loading & demand
72. 72
4.1: Improving the networkes
There are different methods in order to improve the network to increase the
voltages and to put the PF within the range. Which will reducethe losses then the
problems for the consumer will decreaseand the costof KWH will decrease.
These methods are:
1. Increasing the swing bus voltage:
2. Tab changing in the transformer:
3. Adding capacitors:
73. 73
4.1A: Improvement by using taps changing:
In this method change the tap ratio of the transformers to 5%.
Figure4.1: 5% tap changer.
Table4.2A(TAP CHANGE) : summary of total generation, loading & demand
74. 74
Note:
After changing the taps of the transformers thelosses in the network decrease
- The losses before= 0.479MW,
- The losses after=0.498MW,
Table4.2B(TAP CHANGE) : summary of total generation, loading & demand
Note:
After changing the taps of the transformers thelosses in the network decrease
- The losses before= 0.559MW,
- The losses after=0.579MW,
75. 75
Note:
After changing the taps of the transformers thelosses in the network decrease
- The losses before= 0.490MW,
- The losses after=0.507MW,
76. 76
4.1B: Power Factor Improvement
The cosine of angle of phasedisplacement between voltage and currentin an AC
circuit is known as Power Factor.
How to improve the P.F?
Where:
Qc: The reactive power to be compensated by the capacitor.
P: The real power of the load.
θ old: The actual power angle.
θ new: The proposed power angle.
Capacitor Banks:
The important of improvement power factor is by adding shuntcapacitor banks at
the buses at both transmission and distribution levels and loads and there are
more effective to add them in the low level Voltages.
77. 77
Effect of Low Power Factor:
1. Higher Apparent Current.
2. Higher Losses in the Electrical Distribution network
3. Low Voltage in the network
Benefits of Improving Power Factor:
1. Lower Apparent Power.
2. Reduces losses in the transmission line.
3. Improves voltage drop.
4. Avoiding the penalties.
The problem of low power factor:
The low P.F is highly undesirable as it causes an increase in the current, resulting
in additional losses of active power in all the elements of power system from
power station generator down to the utilization devices .In additional to the
losses the low P.F causes penalties.
The following table shows the system of the penalties in our companies:
Power Factor P.F Penalties
P.F≥ 0.92 No Penalties.
0.92>P.F ≥0.8 1% of total bill for each one under 0.92
0.8>P.F≥0.7 1.25%of total bill for each one under 0.92
P.F <0.7 1.5%of total bill for each one under 0.92
Table4.3: The penalties of power factor.
Our aim to improvement the P.F in order to avoid penalties and to reduce the
current flow in the network which reduce the electrical losses in the network.
78. 78
Table4.4A(POWER FACTOR) : summary of total generation, loading & demand
Note:
After adding the capacitor banks the losses in the network decrease
- The losses before= 0.498MW,
- The losses after=0.454MW,.
79. 79
Table4.4B(POWER FACTOR) : summary of total generation, loading & demand
Note:
After adding the capacitor banks the losses in the network decrease
- The losses before= 0.579MW,
- The losses after=0.546MW,.
80. 80
Note:
After adding the capacitor banks the losses in the network decrease
- The losses before= 0.507MW,
- The losses after=0.478MW,.
81. 81
4.1C: OverloadedTransformers Problem
After the improvement of the network in the maximum case there is the problem
of the overloaded transformers. This problemwas solved by changing
transformers wherethetransformers which aresmall and the load on them large
were changed with large highly loaded transformer. this will need to buy new
transformers.
Table 4.5 shows the transformers which are needed to be bought:
Number of transformers KVA
1 400
Table 4.6 summarizes the analysis results after changing transformers
82. 82
Chapter 5: Mechanical design
of the network
Mechanical Design:
Electrical Power can be transmitted or distributed either by means of
underground cables or by overhead lines. The underground cables are rarely used
for power transmission dueto main reasons. Firstly, power is generally
transmitted over long distances to load centers. Obviously, theinstallation costs
for underground transmission willbe very heavy. Secondly, electric power has to
be transmitted at high or medium voltages for economic reasons .Itis very
difficult to provideproper insulation to the cables to withstand such higher
pressures. Therefore, as a rule, power transmission over long distances is carried
out by using over headlines. With growth in power demand and consequentrise
in voltage levels, power transmission by overhead lines has assumed considerable
importance.
An overhead line is subjected to uncertain whether conditions and other external
interferences. This calls for the useof proper mechanical factor safety in order to
ensurethe continuity of operation in the line. In general, the strength of the line
should be such so as to provideagainstthe worstprobableweather conditions .In
this chapter; weshall focus our attention on the various aspects of mechanical
design of over head lines.
83. 83
Main Components of overheadlines:
In general, the main components of overhead lines are:
Conductors:
Which carry electric power fromthe sending end station to receiving end
station
Supports:
Which may be poles or towers and keep the conductors at suitable level
above the ground.
Insulators:
Which are attached to supports and insulate the conductors fromthe
ground.
Cross Arms:
Which providesupportto the insulators.
Miscellaneous Items:
Such as phaseplates, danger plates, lightning arrestors, anticlimbing wires[17]
84. 84
Conductors Material :
The conductor is one of the important items as most of the capital outlay is
invested for it. Therefore, proper choice of material and size of the conductor is of
considerableimportance. The conductor material used for transmission and
distribution of electric power should be having the following properties:
1. High electrical conductivity.
2. High tensile strength in order to withstand mechanical stresses.
3. Low cost so that it can be used for long distances.
4. Low specific gravity so that weight per unit volume is small.
The most commonly used conductor materials for overhead lines are
copper, aluminum, steel-cored aluminum, galvanized steel and cadmium
copper.
The choice of a particular material will depend upon the cost, the required
electrical and mechanical Properties and the local conditions.
Copper:
Copper is an ideal material for overhead lines owing to its high electrical
conductivity and greater tensile strength. Itis always used in the hard
drawn formas stranded conductor.
Although hard drawing decreases the electrical conductivity slightly yet it
increases the tensile strength considerably.
Copper has high currentdensity i.e. the current carrying capacity of copper
per unit of X-sectional area is quite large. This leads to two advantages.
Firstly, smaller X-sectional area of conductor
is required and secondly, the area offered by the conductor to wind loads is
reduced. Moreover, this metal is quite homogeneous, durableand has high
scrap value. There is hardly any doubt that copper is an ideal material for
transmission and distribution of
electric power. However, dueto its higher cost and non availability, it is
rarely used for these purposes.
85. 85
Aluminum:
Aluminum is cheap and light as compared to copper but it has much smaller
Conductivity and tensile strength. The relative comparison of the two
materials is briefed below:
1) The conductivity of aluminum is 60% that of copper. The smaller conductivity
of aluminum means that for any particular transmission efficiency, the X-sectional
area of conductor mustbe larger in aluminum than in copper. For the same
resistance, the diameter of aluminum conductor is about 1·26 times the diameter
of copper conductor.
The increased X-section of aluminum exposes a greater surfaceto wind pressure
and, therefore, supporting towers mustbe designed for greater transverse
strength. This often requires the use of higher towers with consequence of
greater sag.
2) The specific gravity of aluminum (2·71 gm/cc) is lower than that of copper (8·9
gm/cc).
Therefore, an aluminum conductor has almost one-half the weight of equivalent
copper conductor.
For this reason, the supporting structures for aluminumneed not be made so
strong as that of copper conductor.
3) Aluminum conductor being light, is liable to greater swings and hence larger
cross-armsare
required.
4) Due to lower tensile strength and higher co-efficient of linear expansion of
aluminum, the sag is greater in aluminum conductors.
Considering the combined properties of cost, conductivity, tensile strength,
weight etc., aluminum has an edge over copper. Therefore, it is being widely used
as a conductor material. Itis particularly profitable to use aluminum for heavy-
currenttransmission wherethe conductor size is
large and its costforms a major proportion of the total costof complete
installation.
86. 86
Steel cordaluminum:
Due to low tensile strength, aluminum conductors producegreater sag. This
prohibits their usefor larger spans and makes them unsuitable for long distance
transmission.
In order to increase the tensile strength, the aluminum conductor is reinforced
with a core of galvanized steel wires. The compositeconductor thus obtained is
known as steel cored aluminum and is abbreviated as A.C.S.R. (aluminum
conductor steel reinforced).
Steel-cored aluminum conductor consists of central core of galvanized steel wires
surrounded by a number of aluminum strands.
The result of this compositeconductor is that steel core takes greater percentage
of mechanical strength while aluminum strands carry thebulk of current. The
steel cord aluminum conductors havethe following advantages :
1) The reinforcementwith steel increases the tensile strength but at the same
time keeps the composite conductor light. Therefore, steel cored aluminum
conductors will producesmaller sag and hence longer spans can be used.
2) Due to smaller sag with steel cored aluminum conductors, towers of smaller
heights can be used.
Fig 5.1: steel cord aluminum conductor.
Galvanizedsteel:
Steel has very high tensile strength. Therefore, galvanized steel conductors can be
used for extremely long spans or for short line sections exposed to abnormally
high stresses dueto climatic conditions. They have been found very suitable in
ruralareas wherecheapness is the main consideration. Due to poor conductivity
and high resistanceof steel, such conductors arenot suitable for transmitting
87. 87
large power over a long distance. However, they can be used to advantage for
transmitting a small power over a small distance where the size of the copper
conductor desirable fromeconomic considerations would be too small and thus
unsuitable for use becauseof poor mechanical strength.
Cadmium copper:
The conductor material now being employed in certain cases is copper alloyed
with cadmium. An addition of 1% or 2% cadmium to copper increases the tensile
strength by about 50% and the conductivity is only reduced by 15% below that of
pure copper. Therefore, cadmium copper conductor can be useful for
exceptionally long spans. However, dueto high cost of cadmium, such conductors
will be economical only for lines of small X-section i.e., where the cost of
conductor material is comparatively smallcompared with the cost of supports.
Supports :
The supporting structures for overhead line conductors are various types of poles
and towers called line supports In general, the line supports should havethe
following properties:
1. High mechanical strength to withstand the weight of conductors and wind
loads.
2. Light in weight without the loss of mechanical strength.
3. Cheap in costand economical to maintain.
4. Longer life.
5. Easy accessibility of conductors for maintenance.
88. 88
The line supports used for transmission and distribution of electric power are
of various types including wooden poles, steel poles, R.C.C. poles and lattice
steel towers. The choice of supporting structurefor a particular case depends
upon the line span, X-sectionalarea, line voltage, costand local conditions.
Wooden poles:
These are made of seasoned wood and are suitable for lines of moderate X-
sectional area and of relatively shorter spans, say up to 50 meters. Such supports
are cheap, easily available, provideinsulating properties and, therefore, are
widely used for distribution purposes in ruralareas as an economical proposition.
The wooden poles generally tend to rot below
the ground level, causing foundation failure. In order to prevent this, the portion
of the pole below the ground level is impregnated with preservativecompounds
like creosoteoil. Double pole structures of the ‘A’ or ‘H’ type are often used to
obtain a higher transversestrength than could be economically provided by
means of single poles.
The main objections to wooden supports are: (i) tendency to rot below the
ground level (ii) comparatively smaller life (20-25 years) (iii) cannotbe used for
voltages higher than 20 kV (iv) less mechanical strength and (v) requireperiodical
inspection.
Steel poles:
The steel poles are often used as a substitute for wooden poles. They possess
greater mechanical strength, longer life and permit longer spans to be used. Such
poles are generally used for distribution purposes in the cities. This type of
supports need to be galvanized or painted in
order to prolong its life. The steel poles are of three types. (i) Rail poles (ii) tubular
poles and
(iii) Rolled steel joints.
89. 89
RCC poles:
The reinforced concrete poles have become very popular as line supports in
recent years. They havegreater 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 holes in the poles facilitate
the climbing of poles and at the same time reduce the weight of line supports.
The main difficulty with the useof these poles is the high cost of transportowing
to their heavy weight. Therefore, such poles are often manufactured at the site in
order to avoid heavy cost of transportation.
Steel towers:
In practice, wooden, steel and reinforced concrete poles are used for distribution
purposes atlow voltages, say up to 11 kV. However, for long distance
transmission athigher voltage, steel towers are invariably employed. Steel towers
have greater mechanical strength, longer life, can withstand mostsevere climatic
conditions and permit the use of longer spans. The risk of interrupted servicedue
to broken or punctured insulation is considerably reduced owing to longer spans.
Tower footings are usually grounded by driving rods into the earth. This minimizes
the lightning troubles as each tower acts as a lightning conductor.
Insulators :
The overhead lines conductors should besupported on the poles or towers in
such a way that currents fromconductors do not flow to earth through supports.
Line conductors mustbe properly insulated from supports. This is achieved by
securing line conductors to supports with the help of insulators.
The insulators providenecessary insulation between line conductors and supports
and thus prevent any leakage currentfromconductors to earth
90. 90
the insulators should havethe following properties :
1. High mechanical strength in order to withstand conductor load, wind load
etc.
2. High electrical resistanceof insulator material in order to avoid leakage
currents to earth.
3. High relative permittivity of insulator material in order that dielectric
strength is high.
Types of Insulators:
Pin type insulators
The part section of a pin type insulator is shown below as the name suggests,
the pin type insulator is secured to the cross-armon the pole. There is a
grooveon the upper end of the insulator for housing the conductor. The
conductor passes through this grooveand is bound by the annealed wire of
the samematerial as the conductor .Pin type insulators areused for
transmission and distribution of electric power at voltages up to 33 kV. Beyond
operating voltage of 33 kV, the pin type insulators become too bulky and
hence uneconomical. Causes of insulator failure. Insulators arerequired to
withstand both mechanical and electrical stresses. Thelatter type is primarily
due to line voltage and may causethe breakdown of the insulator. The
electrical breakdown of the insulator can occur either by flash-over or
puncture. In flashover, an arc occurs between the line conductor and insulator
pin (i.e. earth) and the dischargejumps across theair gaps, following shortest
distance. In caseof flash-over, theinsulator will continue to act in its proper
capacity unless extreme heat produced by the arc destroys theinsulator. In
case of puncture, the dischargeoccurs fromconductor to pin through the body
of the insulator. When such breakdown is involved, the insulator is
permanently destroyed due to excessiveheat. In practice, sufficient thickness
of porcelain is provided in the insulator to avoid punctureby the line voltage.
The ratio of puncture strength to flashover voltage is known as safety factor.
𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑖𝑠𝑛𝑢𝑙𝑎𝑡𝑜𝑟 =
𝑝𝑢𝑛𝑐𝑡𝑢𝑟𝑒 𝑠𝑡𝑟𝑒𝑛𝑔ℎ
𝑓𝑙𝑎𝑠ℎ − 𝑜𝑣𝑒𝑟 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
91. 91
Fig.5.2: pin type insulators.
Suspensiontype insulators
The cost of pin type insulator increases rapidly as the working voltage is
increased. Therefore, this type of insulator is not economical beyond 33 kV.
For high voltages (>33 kV).
They consistof a number of porcelain discs connected in series by metal links
in the formof a string. The conductor is suspended at the bottom end of this
string while the other end of the string is secured to the cross-armof the
tower. Each unit or disc is designed for low voltage, say 11 kV. The number of
discs in series would obviously depend upon the working voltage. For instance,
if the working voltageis 66 kV, then sixdiscs in series will be provided on the
string.
Advantages:
1) Suspension typeinsulators are cheaper than pin type insulators for
voltages beyond 33 kV.
2) Each unit or disc of suspension type insulator is designed for low voltage,
usually 11 kV.
Depending upon the working voltage, the desired number of discs can be
connected in series.
3) If any one disc is damaged, the whole string does not become useless
because the damaged disc can be replaced by the sound one.
4) The suspension arrangementprovides greater flexibility to the line. The
connection at the cross armis such that insulator string is free to swing in
any direction and can take up the position wheremechanical stresses are
92. 92
minimum.
5) In case of increased demand on the transmission line, it is found more
satisfactory to supply the greater demand by raising the line voltage than to
provideanother set of conductors.
The additional insulation required for the raised voltage can be easily
obtained in the suspension arrangementby adding the desired number of
discs.
6) The suspension typeinsulators aregenerally used with steel towers. As
the conductors run below the earthed cross-armof thetower, therefore,
this arrangementprovides partial protection fromlightning
Fig.5.3: suspension insulators.
Straininsulators
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 relieve the line of excessivetension,
strain insulators are used. For low voltage lines (< 11 kV), shackleinsulators are
used as strain insulators. However, for high voltage transmission lines, strain
insulator consists of an assembly of suspension insulators. Thediscs of strain
insulators are used in the vertical plane. When the tension in lines is
exceedingly high, as at long river spans, two or morestrings are used in
parallel.
93. 93
Fig.5.4: strain insulator.
Important Points:
There is some of criteria wemust take it into account in mechanical design of
medium voltage:
1. Distances between the towers
The distance between each tow towers in the range (80 – 100) meter
2. The high of tower is (12) meter in 33 (KV) voltage level and 22 (KV).
3. Thickness of the steel material (80 – 90) mm, 90 mm fromthe base of tower
and 80 mm toward the top of tower .
4. Base of tower (2-2-2.5) m, (0.5) mabovethe ground and each base need about
(8) cup of concrete.
5. The distances between the insulators in the range (0.5-1) m.
6. Number of steel truss which we can put it in series in straightline without need
to put the steel towers from1 to 4 trusses until about 400 m distance.
The Type of conductor that we usedin this project is Steel CordAluminium.
94. 94
We usedinthis project twotypes of supports, one of themcalledsteel truss
and other type calledsteel towers as we see inthe figure below:
Fig.5.5:truss with arm in front . Fig.5.6:tower.
Total number of steel towers and steel ladders we are need it to connect Tulkarim
and Sarra connection point with each other to unified the whole electrical
network:
Table 5.1: total number of tower and truss we are need to unified tulkarim network and sarra
connection point .
Steel tower Steel truss
61 182
Type of insulator we are used it in this projectis PIN type insulators , Suspension
type insulators and Straininsulators .Tablebelow shows the total number of
insulators weare need it :
95. 95
Number of insulators that we needin eachtype as shown in table below:
Types of insulator Pin Suspension Strain
Number 546 62 366
Table 5.2:total numberwe are needineach type of insulator.
Sag in Overhead Lines:
While erecting an overhead line, it is very important that conductors are under
safetension. If the conductors aretoo much stretched between supports in a bid
to save conductor material, the stress in the conductor may reach unsafevalue
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 a dip or sag. The difference in level between points of supports and the
lowest point on the conductor is called sag.
Figurebelow shows a conductor suspended between two supports A and B. The
conductor is not fully stretched but is allowed to havea dip. The lowestpoint on
the conductor is O and the sag is S.
Fig 5.7: sag in overhead lines.
96. 96
The following points may be noted:
1) When the conductor is suspended between two supports atthe samelevel, it
takes the shapeof catenaries. However, if the sag is very small compared with the
span, then sag-span curveis like a parabola.
2) The tension at any point on the conductor acts tangentially. Thus tension TO at
the lowestpoint O acts horizontally.
3) The horizontalcomponent of tension is constantthroughoutthe length of the
wire.
4) The tension at supports is approximately equal to the horizontaltension acting
at any point on the wire. Thus if T is the tension at the supportB, then T = TO.
Conductor sag and tension. This is an important consideration in the mechanical
design of overhead lines. The conductor sag should be kept to a minimum in
order to reducethe conductor material required and to avoid extra pole height
for sufficient clearance above ground level. Itis also desirable that tension in the
conductor should be low to avoid the mechanical failure of conductor and to
permit the use of less strong supports. However, low conductor tension and
minimum sag are not
possible. Itis becauselow sag means a tight wire and high tension, whereas a low
tension means a loose wire and increased sag. Therefore, in actual practice, a
compromisein made between the two.
Calculation of Sag
In an overhead line, the sag should be so adjusted that tension in the conductors
is within safelimits. The tension is governed by conductor weight, effects of wind,
ice loading and temperature variations.
Itis a standard practice to keep conductor tension less than 50% of its ultimate
tensile strength i.e.. We shall now calculate sag and tension of a conductor when
1) supports areat equal levels and 2) supports areat unequal levels.
97. 97
When supports are at equal levels:
Consider a conductor between two equivalent supports A and B with O as the
lowest point as shown in Fig.
Let Fig.5.8: sag when supports are at equal levels.
l = Length of span
w = Weight per unit length of conductor
T = Tension in the conductor.
we get,
15]
When supports are at unequal levels:
Fig. below shows a conductor suspended between two supports A and B which
are at different levels. The lowestpoint on the conductor is O.
