PDH-Pro.com 396 Washington Street, Suite 159, Wellesley, MA 02481 Telephone β (508) 298-4787 www.PDH-Pro.com This document is the course text. You may review this material at your leisure before or after you purchase the course. In order to obtain credit for this course, complete the following steps: 1) Log in to My Account and purchase the course. If you donβt have an account, go to New User to create an account. 2) After the course has been purchased, review the technical material and then complete the quiz at your convenience. 3) A Certificate of Completion is available once you pass the exam (70% or greater). If a passing grade is not obtained, you may take the quiz as many times as necessary until a passing grade is obtained (up to one year from the purchase date). If you have any questions or technical difficulties, please call (508) 298-4787 or email us at [email protected]. Mechanical Design of Overhead Lines Course Number: EE-08-947 PDH: 8 Approved for: AK, AL, AR, GA, IA, IL, IN, KS, KY, LA, MD, ME, MI, MN, MO, MS, MT, NC, ND, NE, NH, NJ, NM, NV, OH, OK, OR, PA, SC, SD, TN, TX, UT, VA, VT, WI, WV, and WY New Jersey Professional Competency Approval #24GP00025600 North Carolina Approved Sponsor #S-0695 Maryland Approved Provider of Continuing Professional Competency Indiana Continuing Education Provider #CE21800088
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PDH-Pro.com
396 Washington Street, Suite 159, Wellesley, MA 02481 Telephone β (508) 298-4787 www.PDH-Pro.com
This document is the course text. You may review this material at your leisure before or after you purchase the course. In order to obtain credit for this course, complete the following steps: 1) Log in to My Account and purchase the course. If you donβt have an account, go to New User to create an account. 2) After the course has been purchased, review the technical material and then complete the quiz at your convenience. 3) A Certificate of Completion is available once you pass the exam (70% or greater). If a passing grade is not obtained, you may take the quiz as many times as necessary until a passing grade is obtained (up to one year from the purchase date). If you have any questions or technical difficulties, please call (508) 298-4787 or email us at [email protected].
New Jersey Professional Competency Approval #24GP00025600 North Carolina Approved Sponsor #S-0695 Maryland Approved Provider of Continuing Professional Competency Indiana Continuing Education Provider #CE21800088
Mechanical Design of Overhead Lines
Copyright 2018 Velimir Lackovic Page 1
Mechanical Design of Overhead Lines
Electric power can be carried either by underground cables or overhead transmission and
distribution lines. The underground cables are not typically used for power transmission due to
two reasons.
1. Power is carried over long distances to remote load centres. Obviously, the installation
costs for underground transmission will be huge.
2. Electric power has to be transferred at high voltages for economic reasons. It is very difficult
to achieve proper insulation to the cables to withstand higher pressures.
Therefore, power transfer over long distances is done by using overhead lines. With the power
demand increase and consequent voltage level rise, power transmission by overhead lines has
assumed significant importance.
Nevertheless, an overhead line is subjected to various weather conditions and other external
interferences. This asks for the use of adequate mechanical safety factors in order to ensure the
continuity of line operation. Typically, the strength of the line needs to be such so it can
withstand the worst probable weather conditions. This course focuses on the different aspects
of mechanical design of overhead lines.
Overhead Line Main Components
An overhead line may be used to transfer or distribute electric power. The proper overhead line
operation depends to a big extent upon its mechanical design. While constructing an overhead
line, it has to be verified that line mechanical strength is such so as to provide against the most
probable weather conditions. Typically, the main elements of an overhead line are:
- Conductors which transfer power from the sending end station to the receiving end
station.
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- Supports which may be poles or towers. They keep the conductors at an appropriate
level above the earth.
- Insulators that are connected to supports and insulate the conductors from the earth.
- Cross arms which give support to the insulators.
- Miscellaneous elements such as phase plates, danger plates, surge arrestors, etc.
The overhead line operation continuity depends upon the judicious selection of above
elements. Hence, it is beneficial to have detailed discussion on them.
Overhead Line Conductor Materials
The conductor is one of the crucial items as most of the financial outlay is invested for it. Hence,
correct selection of conductor material and size is of significant importance. The conductor
material used for transmission and distribution of electric power needs to have the following
characteristics:
- High tensile strength in order to sustain mechanical stresses
- High electrical conductivity
- Low specific gravity so that weight per unit volume is small
- Low cost so that it can be used for considerable distances
All above demands cannot be found in a single material. Hence, while choosing a conductor
material for a particular application, a compromise is made between the cost and the needed
electrical and mechanical characteristics.
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Typically used conductor materials
Typically used conductor materials for overhead lines are copper, aluminium, steel-cored
aluminium, galvanised steel and cadmium copper. The selection of a particular material is
dependant on the cost, the needed electrical and mechanical characteristics and the local
conditions. All conductors used for overhead lines are typically stranded in order to increase
the flexibility. In stranded conductors, there is typically one central wire and around it,
successive layers of wires containing 6, 12, 18, 24 ...... wires. Therefore, if there are n layers, the
overall number of individual wires is 3n(n+1)+1. In the production process of stranded
conductors, the consecutive layers of wires are twisted or spiralled in different directions so
that layers are bound together.
- Copper. Copper is perfect material for overhead lines owing to its great electrical
conductivity and increased tensile strength. It is typically used in the hard drawn form as
stranded conductor. Even though hard drawing slightly decreases the electrical conductivity, it
considerably increases the tensile strength. Copper has great current density. For example, the
current carrying capacity of copper per unit of cross-sectional area is significant. This leads to
two benefits. Firstly, smaller conductor cross-sectional area is needed and secondly, the area
offered by the conductor to wind loads is decreased. Also, this metal is homogeneous, durable
and has big scrap value. There is no doubt that copper is perfect material for electric power
transmission and distribution. Nevertheless, due to its big cost and non-availability, it is not
often used for these purposes. Current trend is to use aluminium instead of copper.
- Aluminium. Aluminium is cheap and light in comparison to copper but it has
considerably smaller conductivity and tensile strength. The relative comparison of the two
materials is as follows:
o The aluminium conductivity is 60% that of copper. The lower aluminium conductivity
means that for any specific transmission efficiency, the conductor cross-sectional area
must be bigger in aluminium than in copper. For the same resistance, the aluminium
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conductor diameter is around 1.26 times the copper conductor diameter. The increased
aluminium cross-section exposes a bigger surface to wind pressure and, hence,
supporting towers have to be designed for greater transverse strength. Typically, this
requires the use of higher towers with consequence of bigger sag.
o The aluminium specific gravity (2.71 gm/cc) is lower than that of copper (8.9 gm/cc).
Hence, an aluminium conductor has almost one-half the weight of equivalent copper
conductor. Due to this, the supporting structures for aluminium need not be made so
strong as that of copper conductor.
o Aluminium conductor being light is liable to bigger swings and therefore bigger cross-
arms are needed.
o Due to lower tensile strength and bigger co-efficient of linear expansion of aluminium,
the sag is bigger in aluminium conductors.
Considering the overall characteristics that include cost, conductivity, tensile strength, weight
etc., aluminium has an edge over copper. Hence, it is being predominantly used as a conductor
material. It is especially profitable to use aluminium for heavy-current transmission where the
conductor size is big and its cost forms a significant proportion of the total cost of complete
installation.
- Steel-cored aluminium. Due to low tensile strength, aluminium conductors have bigger
sag. This forbids their application for bigger spans and makes them unsuitable for long distance
transmission. In order to improve the tensile strength, the aluminium conductor is
strengthened with a core of galvanised steel wires. The obtained composite conductor is known
as steel-cored aluminium or ACSR (aluminium conductor steel reinforced). Steel-cored
aluminium conductor has galvanised steel central core surrounded by a number of aluminium
strands. Typically, diameter of both steel and aluminium wires is the same. Typically, the cross-
section of the two metals are in the ratio of 1:6 but can be modified to 1:4 in order to get more
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conductor tensile strength. Figure 1. presents steel-cored aluminium conductor having one
steel wire surrounded by six aluminium wires. The result of this composite conductor is that
steel core takes bigger percentage of mechanical strength while aluminium strands transfer the
bulk of current.
The steel-cored aluminium conductors have the following benefits:
Figure 1. Steel-cored aluminium conductor having one steel wire surrounded by six aluminium
wires
o The reinforcement with steel improves the tensile strength but at the same time keeps
the composite conductor light. Hence, steel-cored aluminium conductors will create
smaller sag and therefore longer spans can be used.
o Due to smaller sag with steel-cored aluminium conductors, towers of smaller heights
Voltage across top element, π = ( )( ) Voltage across second element from top, π = π (1 + πΎ) Voltage across third element from top, π = π (1 + 3πΎ + πΎ ) %πππ π π‘ππππ ππππππππππ¦ = π£πππ‘πππ πππππ π π π‘πππππ Γ π£πππ‘πππ πππππ π πππ π ππππππ π‘ π‘π πππππ’ππ‘ππ Γ 100
= π3 Γ π Γ 100
The following points have to be noted from the previous mathematical assessment:
- If for example, K=0.2, then from formula (4), we get, V2=1.2V1 and V3=1.64 V1. This
indicates that disc closest to the conductor has maximum voltage across it. The voltage
across other discs decreasing progressively as the cross-arm in reaches.
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- The higher the value of K (=C1/C), the more unequal is the potential across the discs and
lower is the string efficiency.
- The difference in voltage distribution rises with the bigger number of discs. Hence,
shorter string has bigger efficiency than the larger one.
Ways of Enhancing String Efficiency
It has been previously seen that potential distribution in a string of suspension insulators is not
equal. The maximum voltage appears across the insulator closest to the line conductor and
progressively decreases as the cross arm is reached. If the insulation of the most stressed
insulator (closest to conductor) breaks down or flash over takes occurs, the breakdown of other
elements will happen in succession. Therefore it is important to equalise the potential across
the different string elements in order to improve the string efficiency. The different methods
for this are:
- By using longer cross-arms. The string efficiency is dependent upon the value of K. For
example, ratio of shunt capacitance to mutual capacitance. The lower the value of K, the
higher is the string efficiency and more uniform is the voltage distribution. The value of
K can be decreased by decreasing the shunt capacitance. In order to decrease shunt
capacitance, the distance of conductor from tower needs to be increased for example,
longer cross-arms need to be used. Nevertheless, limitations of tower cost and strength
do not allow the use of very long cross-arms. In reality, K=0.1 is the limit that can be
reached by this method.
- By grading the insulators. In this approach, insulators of various dimensions are selected
in a way that each has a different capacitance. The insulators are capacitance graded for
example, they are assembled in the string in such a way that the top element has the
minimum capacitance, growing progressively as the bottom element (closest to
conductor) is reached. Since voltage is inversely proportional to capacitance, this
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method tends to equalise the potential distribution across the string elements. This
approach has the disadvantage that a big number of different-sized insulators are
needed. Nevertheless, good results can be found by using standard insulators for most
of the string and bigger elements for that close to the line conductor.
- By using a guard ring. The potential across each element in a string can be equalised by
applying a guard ring which is a metal ring electrically linked to the conductor and
surrounding the bottom insulator as presented in the Figure 13. The guard ring
introduces capacitance between metal parts and the line conductor. The guard ring is
contoured in a way that shunt capacitance currents i1, i2 etc. are equal to metal fitting
line capacitance currents i1β, i2β etc. The result is that same charging current I flows
through each string element. Finally, there will be equal potential distribution across the
elements.
Figure 12. Insulator shunt capacitance
Tower
Cross arm
Line
Shunt capacitor
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Figure 13. Guard ring
Important Issues
While solving issues relating to string efficiency, the following points have to be considered and
taken into consideration:
- The maximum voltage appears across the disc closest to the conductor (phase
conductor).
- The voltage across the string is equal to phase voltage. For example, voltage across
string=Voltage between line and ground=Phase Voltage
Figure 14. Equivalent circuit for three string insulators Example 2. A 3-phase transmission overhead line is supported by three disc insulators. The
potentials across top element (close to the tower) and middle unit are 8 kV and 11 kV
respectively. Find (a) the ratio of capacitance between pin and ground to the self-capacitance of
each element (b) The line voltage and (c) string efficiency.
- Solution. The equivalent circuit of string insulators is presented in Figure 14. It is given
that V1=8 kV and V2=11 kV.
KC i1
V2
V1
V C
I2
KC i3
KC i2
V3 C
I3
I1 C
A
B
C
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(i) Let K be the ratio of capacitance between pin and ground to self-capacitance. If C is the self-
capacitance of each element, then capacitance between pin and ground= KC. Using Kirchoffβs
current law to point A,
At point A: πΌ = πΌ + π Or π ππΆ = π ππΆ + π πΎππΆ Or π = π (1 + πΎ) Or πΎ = = = 0.375 At point B: πΌ = πΌ + π Or π ππΆ = π ππΆ + (π + π )πΎππΆ Or π = π + (π + π )πΎ = 11 + (8 + 11) Γ 0.375 = 18.12 ππ Voltage between line and ground = π + π + π = 8 + 11 + 18.12 = 37.12 ππ Line voltage = β3 Γ 37.12 = 64.28 ππ ππ‘ππππ ππππππππππ¦ = ππππ‘πππ πππππ π π π‘ππππππ. ππ πππ π’πππ‘πππ Γ π Γ 100 = 37.123 Γ 18.12 Γ 100 = 68.28%
Example 3. Each line of a 3-phase system is supported by a string of 3 similar insulators. If the
voltage across the line unit is 17.5 kV, find the line to neutral voltage. Assume that the shunt
capacitance between each insulator and ground is 1/8th of the capacitance of the insulator
itself. Also calculate the string efficiency.
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- Solution. Figure 15 presents the equivalent circuit of string insulators. If C is the self-
capacitance of each element, then KC will be the shunt capacitance where K=1/8=0.125.
Voltage across line element, V3=17.5 kV
Figure 15. Equivalent circuit of string insulators
At point A: πΌ = πΌ + π Or π ππΆ = π ππΆ + π πΎππΆ Or π = π (1 + πΎ) = π (1 + 0.125) Or π = 1.125 π
KC i1
V2
V1
V C
I2
KC i3
KC i2
V3 C
I3
I1 C
A
B
C
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At point B: πΌ = πΌ + π Or π ππΆ = π ππΆ + (π + π )πΎππΆ Or π = π + (π + π )πΎ = 1.125 π + (π + 1.125 π ) Γ 0.125 Or π = 1.39 π Voltage across top element, π = . = .. = 12.59 ππ Voltage across middle element π = 1.125 π = 1.125 Γ 12.59 = 14.16 ππ Voltage between line and ground (i.e., line to neutral) π + π + π = 12.59 + 14.16 + 17.5 = 44.25 ππ ππ‘ππππ ππππππππππ¦ = ππππ‘πππ πππππ π π π‘ππππππ. ππ πππ π’πππ‘πππ Γ π Γ 100 = 19.053 Γ 7.4 Γ 100 = 85.8%
ππ‘ππππ ππππππππππ¦ = 44.253 Γ 17.5 Γ 100 = 84 β 28% Example 4. The three bus-bar conductors in an outdoor substation are supported by elements
of post type insulators. Each element consists of a stack of 3 pin type insulators linked one on
the top of the other. The voltage across the lowest insulator is 13.1 kV and that across the next
element is 11 kV. Find the bus-bar voltage of the station.
- Solution. The equivalent circuit of insulators is the presented in Figure 15. It is shown
that V3=13.1kV and V2=11 kV. Let K be the ratio of shunt capacitance to self-capacitance
of each element.
Using Kirchhoffβs current law to points A and B, we can derive the following formulas:
π = π (1 + πΎ)
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Or π = (1)
and π = π + (π + π )πΎ (2)
Replacing the value of π = π /(1 + πΎ) in formula (2), we get, π = π + π1 + πΎ + π πΎ Or π (1 + πΎ) = π (1 + πΎ) + [π + π (1 + πΎ)]πΎ = π [(1 + πΎ) + πΎ + (πΎ + πΎ )] = π (1 + 3πΎ + πΎ ) 13.1(1 + πΎ) = 11[1 + 3πΎ + πΎ ] Or 11πΎ + 19.9πΎ β 2.1 = 0 Solving this formula, we get, K=0.1 π = π1 + πΎ = 111 + 0.1 = 10 ππ Voltage between line and ground = π + π + π = 10 + 11 + 13.1 = 34.1 ππ Voltage between bus-bars (i.e., line voltage) = 34.1 Γ β3 = 59 ππ
Example 5. An insulator string has three elements, each having a safe working voltage of 15 kV.
The ratio of self-capacitance to shunt capacitance of each element is 8:1. Calculate the stringβs
maximum safe working voltage. Also calculate the string efficiency.
- Solution. The equivalent circuit of string insulators is presented in Figure 15. The
maximum voltage appears across the lowest unit in the string.
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V3=15 kV; K=1/8=0.125
Using Kirchhoffβs current law at point A, we get,
V2=V1(1+K)
or V1=V2/(1+K)=V2/(1+0.125)=0.89V2 ...(1)
Using Kirchhoffβs current law at point B, we get,
V3=V2+(V1+V2)K=V2+(0.89V2+V2)Γ0.125
π = 1.236 π Voltage across middle element, π = . = . = 12.13 ππ Voltage across top element, π = 0.89π = 0.89 Γ 12.13 = 10.79 ππ Voltage across the string, = π + π + π = 10.79 + 12.13 + 15 = 37.92 ππ String efficiency, = .Γ Γ 100 = 84.26% Example 6. A string of 4 insulators has a self-capacitance equal to 10 times the pin to ground
capacitance. Calculate (a) the voltage distribution across different elements expressed as a
percentage of overall voltage across the string and (b) string efficiency.
- Solution. When the number of insulators in a string surpasses 3, the nodal equation
method becomes tedious. In those cases, there is a simple way to solve the problem. In
this method, shunt capacitance (C1) and self-capacitance (C) of each insulator are
modelled by their equivalent reactances. As it is only the ratio of capacitances which
defines the voltage distribution, the problem can be simplified by assigning unity value
Hence, current through it is K ampere. Therefore, current through second insulator from top is
(1+K) A and voltage across it is (1+K)Γ1=(1+K) V.
With reference to Figure 18 (b), it can be concluded that: ππ = 1 + π1
Or π = π (1 + πΎ) (1) Also ππ = (1 + 3πΎ + πΎ )1
π = π (1 + 3πΎ + πΎ ) (2)
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Dividing (2) by (1) it can be concluded: ππ = 1 + 3πΎ + πΎ1 + πΎ
It is given that π = 18 ππ and π = 13.2 ππ 1813.2 = 1 + 3πΎ + πΎ1 + πΎ Or 13.2πΎ + 21.6πΎ β 4.8 = 0 Solving this formula, we get, K=0.2.
Voltage between line and ground (for example, line voltage) = π + π + π + π = 11 + 13.2 + 18 + 26.49 = 68.69 ππ Voltage between conductors (for example, line voltage) 68.69 Γ β3 = 119 ππ
Example 9. A string of four insulators has a self-capacitance equal to 5 times pin to ground
capacitance. Calculate (a) the voltage distribution across different elements as a percentage of
overall voltage across the string and (b) string efficiency.
- Solution. The ratio of self-capacitance (C) to pin-ground capacitance (C1) is C/C1=5.
The voltage found across the string is expressed as: π = 1 + 1.2 + 1.64 + 2.408 = 6.248 π£πππ‘π The voltage across each element expressed as a percentage of V (for example, 6.248 V) is
expressed by:
Top element = . Γ 100 = 16% Second from top, = .. Γ 100 = 19.2% Third from top, = .. Γ 100 = 26.3% Fourth from top, = .. Γ 100 = 38.5% ππ‘ππππ ππππππππππ¦ = 6.2484 Γ 2.408 Γ 100 = 64.86%
Example 10. The self-capacitance of each element in a string of three suspension insulators is C.
The shunting capacitance of the connecting metal work of each insulator to ground is 0.15 C
while for line it is 0.1 C. Find (a) the voltage across each insulator as a percentage of the line
voltage to ground and (b) string efficiency.
- Solution. In an actual string of insulators, three capacitances exist, self-capacitance of
each insulator, shunt capacitance and capacitance of each element to line as presented
in Figure 20 (a). Nevertheless, capacitance of each element to line is very small and is
typically neglected. Figure 20 (b) presents the equivalent circuit of string insulators.
- Or 0.1π = 1.15π β 1.1π - - Or π = 11.5 π β 11π (1)
Figure 20. Equivalent circuit of string insulators At point B: πΌ + π = πΌ + π Or π ππΆ + π Γ 0.1πΆ Γ π = π ππΆ + (π + π )π Γ 0.15πΆ Or 1.1π = 1.15π + 0.15π Replacing the value of π from expression (a) into expression (b), we get, = 1.1(11.5 π β 11π ) = 1.15π + 0.15π (2)
0.1C
Tower
0.15C
Line
0.15C
0.15C
0.1C
Line capacitance
(a)
V2
V1
V3
0.15C
0.15C
0.15C
C
C
C
(b)
0.1C
0.1C
A
B
I1
i1
i2
i3
I3
I2
i2β
i1β
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Or 13.25π = 12.5π Or π = .. π (3)
Replacing the value of π from expression (3) into expression (1), it can be found: π = 11.5π β 11 12.5π13.25 = 14.813.25 π Now voltage between conductor and ground is: π = π + π + π = π 1 + 12.513.25 + 14.813.25 = 40.55π13.25 π£πππ‘π
π = 13.25π40.55 = 0.326 π π = 12.5 Γ 0.326π13.25 = 0.307 π π = 14.8 Γ 0.326π13.25 = 0.364 π (1) The voltage across each element determined as a percentage of V becomes: Top element, = π Γ = 0.326 Γ 100 = 32.6% Second from top, = π Γ = 0.307 Γ 100 = 30.7% Third from top, = π Γ = 0.364 Γ 100 = 36.4% ππ‘ππππ ππππππππππ¦ = π3 Γ 0.364π Γ 100 = 91.5% Example 11. Each phase of a 3-phase system is suspended by a string of 3 same insulators of
self-capacitance C. The shunt capacitance of connecting metal work of each insulator is 0.2 C to
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ground and 0.1 C to line. Find the system string efficiency if a guard ring increases the
capacitance to the line of metal work of the lowest insulator to 0.3 C.
- Solution. The capacitance between each element and phase is artificially increased by
applying a guard ring as presented in Figure 21. This configuration tends to equalise the
potential across different elements and therefore leads to enhanced string efficiency. It
is given that with the use of guard ring, insulator link-pin capacitance to the line of the
lowest element is increased from 0.1 C to 0.3 C.
Figure 21. Equivalent circuit of string insulators At point A:
At point B: πΌ + π = πΌ + π Or π ππΆ + π Γ 0.3πΆ Γ π = π ππΆ + (π + π )π Γ 0.2πΆ Or 1.3π = 1.2π + 0.2π Replacing the value of π from expression (1) into expression (2), we find, 1.3(12π β 11π ) = 1.2π + 0.2π Or 15.5π = 15.4π π = 15.4π15.5 = 0.993π Replacing the value of π from expression (3) into expression (1), we get,
π = 12π β 11 Γ 0.993π = 1.077π Voltage between conductor and ground (for example, line voltage) = π + π + π = π + 0.993π + 1.077π = 3.07π ππ‘ππππ ππππππππππ¦ = 3.07π3 Γ 1.077π Γ 100 = 95%
Example 12. It is needed to grade a string having seven suspension insulators. If the pin to
ground capacitance are all same to C, find the phase to pin capacitance that would give the
same voltage across each string insulator.
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- Solution. Let C1, C2... C6 respectively be the required line to pin capacitances of the
elements as presented in Figure 22. As the voltage across each insulator has to be the
same:
I1=I2=I3=I4=I5=I6=I7
At point A: πΌ + π = πΌ + π Or π = π (πΌ = πΌ )
As π, πΏ, π, π are the same for the two cases, πβ(π β π ) For first case, P=53 kW and π = β = 61.2 ππ For second case, P=98 kW and π = .β = 64 ππ 53β(61.2 β π ) 98β(64 β π ) Dividing the last two formulas, it can be found: 9853 = (64 β π )(61.2 β π )
π = 54 ππ Let W kilowatt be the power loss at 113 kV πβ 113β3 β π
β(65.2 β 54) Dividing the last two formulas, it can be found:
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π53 = (65.2 β 54)(61.2 β 54)
π = 11.27.2 Γ 53 = 128 ππ
Overhead Line Sag
While building an overhead line, it is crucial that conductors are under safe tension. If the
conductors are too stretched between supports in an attempt to save conductor material, the
stress in the conductor may reach critical value and in some cases the conductor may break due
to excessive tension. In order to secure conductor safe tension, they are not completely
stretched but are allowed to have a dip or sag. The difference in level between support points
and the conductor lowest point is called sag. Figure 23 (a) presents a conductor suspended
between two equilevel supports A and B. The conductor is not completely stretched but is
allowed to have a dip. The conductor lowest point is O and the sag is S. The following items can
be noted:
Figure 23. Conductor suspension between two supports
- When the conductor is suspended between two supports at the same level, it forms the
shape of catenary. Nevertheless, if the sag is very small in comparison with the span,
then sag-span curve is like a parabola.
- The tension at any point on the conductor acts tangentially. Therefore, tension TO at the
lowest point O acts horizontally as presented in Figure 23 (b).
S
A B
O (a)
O
B T
(b) T0
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- The horizontal tension component is constant throughout the wire length.
- The tension at supports is roughly equal to the horizontal tension acting at any point on
the wire. Therefore, if T is the tension at the support B, then T=TO.
Sag and tension of the conductor
This is an important point in the overhead line mechanical design. The conductor sag needs to
be maintained to a minimum in order to decrease the required conductor material and to avoid
extra pole height for sufficient clearance above earth level. It is also preferable that conductor
tension is low to avoid the conductor mechanical failure and to allow the use of less strong
supports. Nevertheless, low conductor tension and minimum sag cannot be achieved. It is
because low sag means a tight wire and high tension, whereas a low tension means a loose wire
and increased sag. Hence in reality, a compromise in made between the two.
Sag Calculation
In an overhead line, the sag has to be adjusted so that tension in the conductors is within safe
boundaries. The tension is governed by conductor weight, wind effects, ice loading and
temperature changes. It is a common practice to maintain conductor tension less than 50% of
its ultimate tensile strength. For example, minimum safety factor in respect of conductor
tension needs to be 2. We shall now find sag and conductor tension when (a) supports are at
equal levels and (b) supports are at different levels.
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Figure 24. Conductor between two equilevel supports
- When supports are at same levels. Consider a conductor between two equilevel
supports A and B with O as the lowest point as presented in Figure 24. It can be shown
that lowest point will be at the mid-span. Consider:
l=Span length
w=Weight per conductor unit length
T=Tension in the conductor.
Consider a point P on the conductor. Considering the lowest point O as the origin, let the co-
ordinates of point P be x and y. Assuming that the curvature is so small that curved length is
equal to its horizontal projection (for example, OP=x), the two forces acting on the portion OP
of the conductor are:
(a) The conductor weight wx acting at a distance x/2 from O.
(b) The tension T acting at O.
Equating the moments of above two forces about point O, we find:
A l/2 l/2 B
T O
X/2 S
y
w.x
X
P
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ππ¦ = π€π₯ Γ π₯2 π¦ = π€π₯2π The maximum dip (sag) is expressed by the value of y at either of the supports A and B. At
support A, x=l/2 and y=S
Sag, π = ( / ) =
- When supports are at different levels. In hilly locations, we typically come across
conductors suspended between supports at different levels. Figure 25 presents a conductor
suspended between two supports A and B which are at different levels. The lowest point on the
conductor is O.
Suppose:
l=Span length
h=Difference in levels between two supports
x1=Distance of support at lower level (for example, A) from O
x2=Distance of support at higher level (for example, B) from O
T=Conductor tension
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Figure 25. Conductor suspended between two different levels If w is the conductor weight per unit length, then,
Solving expressions (1) and (2), it can be found: π₯ = π2 β πβπ€π π₯ = π2 + πβπ€π
Having determined x1 and x2, values of S1 and S2 can be easily found.
Wind and ice loading effect
The above equations for sag are correct only in still air and at normal temperature when the
conductor is acted only by its weight only. Nevertheless, in real life a conductor may have ice
coating and simultaneously exposed to wind pressure. The weight of ice acts vertically
downwards for example, in the same direction as the conductor weight. The force due to the
wind is assumed to act horizontally for example, at right angle to the conductor projected
surface. Therefore, the complete force on the conductor is the vector sum of horizontal and
vertical forces as presented in Figure 26 (c).
Figure 26. Wind effect on the conductor Overall weight of conductor per unit length is:
π€ = (π€ + π€ ) + (π€ ) Where w - conductor weight per unit length (conductor material density x volume per unit length)
Wind
(b)
WW
Wt (W+ Wi)
(c)
ΞΈ
(a)
Ice coating
t d
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π€ - ice weight per unit length (density of ice x volume of ice per unit length) w - wind force per unit length (wind pressure per unit area x projected area per unit length)
When the conductor has wind and ice loading, the following points have to be considered:
- The conductor sets itself in a plane at an angle to the vertical where
tan π = π€π€ + π€
- The sag in the conductor is expressed as:
π = π€ π2π Therefore, S represents the slant sag in a direction making an angle to the vertical. If no specific
mention is made in the problem, then slant slag is found by using the above equation.
- The vertical sag=ScosΞΈ
Example 17. A 132 kV transmission line has the following parameters:
Weight of conductor=680 kg/km; Length of span=260 m
Overall strength=3100 kg; Safety factor=2
Find the height above earth at which the conductor needs to be supported. Required earth
clearance is 10 metres.
- Solution.
Weight of conductor/metre run, w = 680/1000 = 0.68 kg
ππππ πππππ‘β, π = 260 π πππ = π€π8π = 0.68 Γ 2608 Γ 1550 = 3.7 π Conductor has to be supported at a height of 10+3.7=13.7 m Example 18. A transmission line has a span of 150 m between level supports. The conductor
has a cross-sectional area of 2 cm2. The tension in the conductor is
2000 kg. If the conductor material specific gravity is 9.9 gm/cm3 and wind pressure is 1.5 kg/m
length, find the sag. Calculate also the vertical sag.
- Solution.
Span length, l=150 m; Working tension, T=2000 kg
Wind force/m length of conductor, ww=1.5 kg
Weight of conductor/m length, w=Specific Gravity Γ Volume of 1 m conductor
Figure 27. Overall conductor weight This is the value of slant sag in a direction making an angle with the vertical. Going back to
Figure 27, the value of ΞΈ is expressed as:
tan π = π€π€ = 1.51.98 = 0.76
π = π‘ππ 0.76 = 37.23Β° Vertical sag = π cos π = 3.48 Γ cos 37.23Β° = 2.77 π Example 19. A transmission line has a span of 200 metres between level supports. The
conductor has a cross-sectional area of 1.29 cm2, weighs 1170 kg/km and has a breaking stress
of 4218 kg/cm2. Find the sag for a safety factor of 5, allowing a wind pressure of 122 kg per
square metre of projected surface. Calculate the vertical sag.
= 122 Γ (1.28 Γ 10 Γ 1) = 1.56 ππ Overall weight of conductor per metre length is π€ = π€ + π€ = 1.17 + 1.56 = 1.95 ππ Total weight of conductor per metre length is π€ = π€ + π€ = 1.17 + 1.56 = 1.95 ππ Slant sag, π = = . ΓΓ = 8.96 π The slant sag makes an angle ΞΈ with the vertical where value of ΞΈ expressed as: π = π‘ππ π€π€ = π‘ππ 1.561.17 = 53.13Β° ππππ‘ππππ π ππ = π cos π = 8.96 Γ cos 53.13Β° = 5.37 π Example 20. A transmission line has a span of 275 m between level supports. The conductor
has an effective diameter of 1.96 cm and weighs 0.865 kg/m. Its overall strength is 8060 kg. If
the conductor has ice coating of radial thickness 1.27 cm and is exposed to a wind pressure of
3.9 gm/cm2 of projected surface, compute sag for a safety factor of 2. Weight of 1 c.c. of ice is
0.91 gm.
- Solution.
Span length, l=275 m ; Weight of conductor/m length, w=0.865 kg
Conductor diameter, d=1.96 cm; Ice coating thickness, t=1.27 cm Working tension, T
=8060/2=4030 kg Volume of ice per metre (for example, 100 cm) length of conductor
= ππ‘(π + π‘) Γ 100ππ = π Γ 1.27 Γ (1.96 + 1.27) Γ 100 = 1288 ππ Weight of ice per metre length of conductor is:
Overall weight of one metre length of conductor is π€ = π€ + π€ = 1.78 + 1.5 = 2.33 ππ Slant sag, π = = . ΓΓ = 3.28 π ππππ‘ππππ π ππ = π cos π = 3.28 Γ π€π€ = 3.28 Γ 1.782.33 = 2.5 π
Conductor has to be supported at a height of 7+2.5=9.5 m
Example 23. The towers of height 30 m and 90 m respectively support a overhead line
conductor at water crossing. The horizontal distance between the towers is 500 m. If the
tension in the conductor is 1600 kg, calculate the minimum clearance of the conductor and
water and clearance mid-way between the supports. Weight of conductor is 1.5 kg/m. Bases of
the towers can be considered to be at water level.
Figure 28 presents the conductor suspended between two supports A and B at different levels
with O as the lowest point on the conductor. Here, l=500 m; w=1.5 kg; T=1600 kg. Difference in
support levels, h=90β30=60 m. Let the lowest point O of the conductor be at a distance x1 from
the support at lower level (for example, support A) and at a distance x2 from the support at
higher level (for example, support B).
Apparently, x1+x2=500 m
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Figure 28. Conductor suspended between two different levels Now π = and π = β = π β π = π€π₯2π β π€π₯2π Or 60 = π€2π (π₯ + π₯ )(π₯ β π₯ ) π₯ β π₯ = 60 Γ 2 Γ 16001.5 Γ 500 = 256 π Solving above formulas it can be found: π₯ = 122 π, π₯ = 378 π Now, π = = . ΓΓ = 7 π Lowest point O clearance from water level=30-7=23 m Let the mid-point P be at a distance x from the lowest point O. Apparently, π₯ = 250 β π₯ = 250 β 122 = 128 π Sag at mid-point P, π = = . ΓΓ = 7.68 π Clearance of mid-point P from water level=23+7.68=30.68 m
B
S1
S2
h
P
O
A
x
x1 x2
l=500 m
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Example 24. An overhead transmission line conductor with a parabolic arrangement weighs
1.925 kg per metre of length. The area of cross-section of the conductor is 2.2 cm2 and the
overall strength is 8000 kg/cm2. The supports are 600 m apart having 15 m difference of levels.
Find the sag from the taller of the two supports which have to be allowed so that the safety
factor shall be 5. Consider that ice load is 1 kg per metre run and there is no wind pressure.
- Solution. Figure 29 presents the conductor suspended between two supports at A and B
at different levels with O as the lowest point on the conductor.
Here, l=600 m; wi=1 kg; h=15m
w=1.925 kg; T=8000Γ2.2/5=3520 kg
Complete weight of 1 m length of conductor is:
wt=w+wi=1.925+1=2.925 kg
Let the conductor lowest point O be at a distance x1 from the support at lower level (for
example, A) and at a distance x2 from the support at higher level (for example, B).
Example 25. An overhead line at a river crossing is supported from two towers at heights of 40
m and 90 m above water level, the horizontal distance between the towers being 400 m. If the
maximum allowable tension is 2000 kg, calculate the clearance between the conductor and
water at a point mid-way between the towers. Conductor weight is 1 kg/m.
- Solution. Figure 30 shows the complete configuration.
O
A
B
S1
S2
x1 x2
600 m
h
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Figure 30. Overhead line at a river crossing Here, h=90β40=50 m; l=400 m
T=2000 kg; w=1 kg/m
Apparently, x1+x2=400 m (1)
Now: β = π β π = π€π₯2π β π€π₯2π Or 50 = π€2π (π₯ + π₯ )(π₯ β π₯ ) π₯ β π₯ = Γ Γ = 500 π (2) Solving expressions (1) and (2), it can be found, x2=450 m and x1=β50 m
Now x2 is the distance of higher support B from the conductor lowest point O, whereas x1 is
that of lower support A. As the span is 400 m, hence, point A lies on the same side of O as B (as
presented in Figure 30). Horizontal distance of mid-point P from lowest point O is
x
B
P
A O
S2
Smid
S1
X1=50m
X2=450m
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x=Distance of A from O+400/2=50+200=250 m
Sag at point P, π = = ΓΓ = 15.6 π Now Sag π = = ΓΓ = 50.6 π Height of point B above mid-point P: = π β π = 50.6 β 15.6 = 35 π Clearance of mid-point P above water level=90-35=55 m Example 26. An overhead line over a hillside with the gradient of 1:20, is supported by two 22
m high towers with a distance of 300 m between them. The lowest conductor is fixed 2 m
below the top of each tower. Calculate the clearance of the conductor from the earth. Given
that conductor weighs 1 kg/m and the allowable tension is 1500 kg.
- Solution. The conductors are supported between towers AD and BE over a hillside
having gradient of 1:20 as presented in Figure 31. The lowest point on the conductor is
O and sinΞΈ=1/20.
Effective height of each tower (AD or BE)=22β2=20 m
Clearance of the lowest point O from the earth is: ππΊ = π»πΉ β π β πΊπΉ = π΅πΆ β π β πΊπΉ Now πΊπΉ = π₯ tan π = 75 Γ 0.05 = 3.75 π = 35 β 16.87 β 3.75 = 14.38 π
O
A
B
S1
S2
x1 x2
G
D F C
E
ΞΈ
H
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Example 27. A transmission tower on a level earth gives a minimum clearance of 8 metres for
its lowest conductor with sag of 10 m for a span of 300 m. If the same tower is to be used over
a slope of 1 in 15, calculate the minimum earth clearance obtained for the same span, same
Figure 32. The conductors supported between towers over a sloping ground
Solving formulas (1) and (2), it can be found x1=75 m and x2=225 m
Now π = π€π₯2π = 8 Γ 752 Γ 9 Γ 10 = 2.5 π π = π€π₯2π = 8 Γ 2252 Γ 9 Γ 10 = 22.5 π Point O clearance from the earth is:
OG=BCβS2βGF=38β22.5β5=10.5 m
GF=x1 tanΞΈ=75 Γ 1/15 = 5m]
Since O is the origin, the equation of slope of earth is expressed as: π¦ = ππ₯ + π΄ Here: π = 115 πππ π΄ = ππΊ = β10.5 π
O
A
B
S1
S2
x1 x2
G
D F C
E
ΞΈ
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π¦ = π₯15 β 10.5 Clearance C from the earth at any point x is: πΆ = πΈππ’ππ‘πππ ππ πππππ’ππ‘ππ ππ’ππ£π β π¦ = π€π₯2π β π₯15 β 10.5
Clearance will be minimum when = 0 for example, πππ₯ π₯2250 β π₯15 + 10.5 = 0
Or 2π₯2250 β 115 = 0 Or π₯ = 115 Γ 22502 = 75 π For example, minimum clearance will be at a point 75 m from O. ππππππ’π πππππππππ = π₯2250 β π₯15 + 10.5 = 752250 β 7515 + 10.5 = 2.5 β 5 + 10.5 = 8 π
Certain Mechanical Principles
Mechanical safety factors to be used in overhead line design should depend to some degree on
the continuity of operation importance. In principle, the line strength needs to be such as to
provide against the worst expected weather conditions. Some important issues in the overhead
transmission lines mechanical design are:
- Tower height: Tower height is dependent upon the span length. With long spans,
relatively few towers are needed but they have to be tall and correspondingly costly.
Typically, it is not possible to find the tower height and span length on the basis of direct
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construction costs because the lightning hazards highly increase as the height of the
conductors above earth is increased. This is one reason that horizontal spacing is
preferred in spite of the wider right of way.
- Conductor clearance to earth: The conductor clearance to earth at the time of biggest
sag should not be less than some specified distance (typically between 6 and 12 m),
depending on the voltage, on the nature of the country and on the local regulations. The
biggest sag may happen on the hottest day of summer on account of the expansion of
the wire or it may happen in winter owing to the formation of a heavy coating of ice on
the wires. Specific provisions must be made for melting ice from the power lines.
- Sag and tension: When installing overhead transmission lines, it is mandatory to allow a
reasonable safety factor in respect of the tension to which the conductor is subjected.
The tension is governed by the wind effects, ice loading and temperature changes. The
relationship between tension and sag depends on the loading conditions and
temperature changes. For instance, the tension increases when the temperature
reduces and there is a corresponding sag reduction. Icing-up of the line and wind
loading will cause conductor stretching by an amount dependent on the line tension.
In sag planning, tension and clearance to earth of a given span, a maximum stress is chosen. It is
then aimed to have this stress developed at the worst probable weather conditions (for
example, minimum expected temperature, maximum ice loading and maximum wind). Wind
loading increases the sag in the direction of resultant loading but reduces the vertical
component. Hence, in clearance calculations, the wind effect should not be included unless
horizontal clearance is important.
- Stringing charts: For use in the field work of stringing the conductors, temperature-sag
and temperature tension charts are printed for the given conductor and loading conditions.
Such curves are known as stringing charts and are shown in Figure 33. These charts are very
useful while stringing overhead transmission lines.
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- Conductor spacing: Conductor spacing has to be such to provide safety against flash-
over when the wires are swinging in the wind. The adequate spacing is a function of span
length, voltage and weather conditions. The use of horizontal spacing removes the danger
caused by unequal ice loading. Small wires or light material wires are exposed to more swinging
by the wind than heavy conductors. Hence, light wires need to be provided bigger spacings.
- Conductor vibration: Wind exerts pressure on the exposed conductor area. If the wind
velocity is small, the conductor swinging is harmless given the clearance is sufficiently big so
that conductors do not approach within the sparking distance of each other. A totally different
vibration type, known as dancing, is caused by the action of fairly strong wind on a wire covered
with ice, when the ice coating happens to take a form which makes a good air-foil section. Then
the complete span may sail up like a kite until it reaches the limit of its slack, stops with a jerk
and falls or sails back. The harmful impact of these vibrations happens at the clamps or
supports where the conductor suffers fatigue and finally breaks. In order to save the