tag:blogger.com,1999:blog-36914023977208519762024-03-29T00:09:35.888-07:00Electrician TheoryUnknownnoreply@blogger.comBlogger84125tag:blogger.com,1999:blog-3691402397720851976.post-59030927364116246762013-04-21T06:35:00.001-07:002013-04-21T06:35:19.598-07:00Testing of Cables The cables are tested both before and after installation.<br />
Before installation the cables are tested for the following ;<br />
1. Continuity test 2. Insulation test<br />
<div style="text-align: justify;">
Both the tests can be performed
with the help of megger. The values of inductance, capacitance and
resistance of the cable are also recorded before installation. These
values help in locating the faults in future when the cable is
installed.</div>
<div style="text-align: justify;">
When there are
faults in the installed cable then quick identification of location of
the fault is very much necessary for quick repairs. For this the
following tests are conducted on the cables,</div>
1. Murray loop test<br />
2. Fall of potential test<br />
3. D.C. charge and discharge test.<br />
<div style="color: blue;">
<u>1.1 Murray Loop Test</u></div>
<div style="text-align: justify;">
In this test, the principle of wheatstone's bridge is used to locate
the ground faults. In ground fault, one or more cable cores get earthed.
This is the most accurate test.</div>
<div style="text-align: justify;">
In this test, one sound return wire of the same cross-section as that
of cable is required. The connections are shown in the Fig. 1.</div>
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhnn18JTRkf70WQwZjnx1RXLtatL7yfgUyHEMUENnT0ntcDMOcJ__amMOMIC9W5VV-ypZ3SlJZA0go44dhRcC-di6YgTgllJ5RMvbVOVTGrMnZ-KmniH_X1FZe2GhUpO35f75WCH-es-Zs/s1600/ABB1.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhnn18JTRkf70WQwZjnx1RXLtatL7yfgUyHEMUENnT0ntcDMOcJ__amMOMIC9W5VV-ypZ3SlJZA0go44dhRcC-di6YgTgllJ5RMvbVOVTGrMnZ-KmniH_X1FZe2GhUpO35f75WCH-es-Zs/s1600/ABB1.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 1</b></td></tr>
</tbody></table>
<div style="text-align: justify;">
The perfectly sound cable is looped with faulty cable. The balance of the bridge is obtained by varying the resistances.</div>
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<div style="text-align: justify;">
<br />
</div>
Where a = Resistance connected to faulty cable<br />
b = Resistance connected to sound cable<br />
Loop length = x + y i.e. 2 times the route length<br />
<div style="color: blue;">
<u>1.2 Fall of Potential Test</u></div>
The arrangement used for the test is shown in the Fig. 2.<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiLx7ckglkL0SzViVIQeJ5miLVTw1tUAWZI532sYMOXyaFZDnIfPn2KfTTP6jKvOm8nAW_FcJDYUtwqVl5ceRINGoYZtg7BDAmZzq35EKaFSJFwEoxprLl3mMZ9oISAmFoJAgyLg_GqR6Q/s1600/ABB12.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiLx7ckglkL0SzViVIQeJ5miLVTw1tUAWZI532sYMOXyaFZDnIfPn2KfTTP6jKvOm8nAW_FcJDYUtwqVl5ceRINGoYZtg7BDAmZzq35EKaFSJFwEoxprLl3mMZ9oISAmFoJAgyLg_GqR6Q/s1600/ABB12.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 2 Fall of potential test</b></td></tr>
</tbody></table>
<div style="text-align: justify;">
The voltmeter
used is high resistance low reading type. The battery, variable
resistance and ammeter are in series. Once the reading of voltmeter is
taken by making connection to the faulty cable and is obtained as V<sub>1</sub>. Other reading is taken by making connection to the second cable and is obtained as V<sub>1</sub>.</div>
Then the fault distance d is given by,<br />
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<br />
Where L = Total equivalent length of the loop<br />
It is necessary to adjust the ammeter reading to the same value for both the readings.<br />
<div style="color: blue;">
<u>1.3 D.C. Charge and Discharge Test</u></div>
This test is used to locate discontinuity in the core of the cable, with high resistance to earth.<br />
The connections are shown in the Fig. 3.<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiDDpnW190mCSAadXLSnaajiXjOSbP5WcXu5kef-F3rMA4TT7bGFd3D6U-sLw6mZBekZZkoNjaY2P5Qg_4Mz1QWgklfbki1A6bjqEUsTq9lqEdRuQnz4z6Lwa2dgMVyvx0ZktKux2Vkl_o/s1600/ABB14.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiDDpnW190mCSAadXLSnaajiXjOSbP5WcXu5kef-F3rMA4TT7bGFd3D6U-sLw6mZBekZZkoNjaY2P5Qg_4Mz1QWgklfbki1A6bjqEUsTq9lqEdRuQnz4z6Lwa2dgMVyvx0ZktKux2Vkl_o/s1600/ABB14.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 3 Charge and discharge test</b></td></tr>
</tbody></table>
The two way switch S is used along with the battery and the galvanometer G.<br />
<div style="text-align: justify;">
The switch position is selected
as 1 and cable is charged first with the help of battery for 15
seconds. Then it is discharged through the galvanometer with switch
position as 2. The galvanometer deflection is observed. Similar readings
are taken at the other end of the cable. Then the distance d of the
fault can be obtained from the end A as,</div>
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<div style="text-align: justify;">
<br />
</div>
<div style="text-align: justify;">
In this test, it is
necessary to earth all the broken cores at far end and and all cores at
far end and all cores except core under test at the test point, to avoid
any false reading.</div>
<div style="text-align: justify;">
Nowadays electronic cable fault locates are available which give the
reading directly on a scale. The principle used in such instruments is
impressing voltage impulses on the cable under test. These impulses get
reflected from the fault location. Then reflections are projected on
C.R.O. in the image from. From this, the distance and type of the cable
fault is determined.</div>
<b><a href="http://www.box.net/shared/qodo8f35n300tp4fy3j7">Read solved examples</a></b>Unknownnoreply@blogger.com7tag:blogger.com,1999:blog-3691402397720851976.post-37797870235085373932013-04-21T06:33:00.001-07:002013-04-21T06:33:12.264-07:00Current Rating of Cable Once all the thermal
resistances are known then the current carrying capacity of a cable can
be determined. When the cables carry an excessive current the cable get
heated up. It is not advisable to operate the cable at excessively high
temperature because of the following reasons :<br />
<div dir="ltr" style="text-align: left;">
1. Due to high temperature, oil in the oil filled cables get expanded and this may lead to bursting of the sheath.</div>
<div dir="ltr" style="text-align: justify;">
2.
High temperature can cause unequal expansion which leads to the
information of voids. Such voids may lead to ionisation and finally lead
to insulation failure.</div>
<div dir="ltr" style="text-align: left;">
3. The dielectric losses increase with temperature which also can lead to breakdown of insulation.</div>
<div dir="ltr" style="text-align: justify;">
Hence the cable must be operated at a current less than the maximum current carrying capacity of the cable.</div>
<div dir="ltr" style="text-align: left;">
The current rating of a cable is dependent on the following factors,</div>
<div dir="ltr" style="text-align: left;">
1. The maximum permissible temperature at which conductor insulation can be operated.</div>
<div dir="ltr" style="text-align: left;">
2. Heat dissipation arrangement through the cable.</div>
<div dir="ltr" style="text-align: left;">
3. The ambient conditions as well as conditions at the time of installation.</div>
<div dir="ltr" style="text-align: left;">
Let us see how to determine current carrying capacity of a cable.</div>
<div dir="ltr" style="text-align: left;">
Let n = Number of phases</div>
<div dir="ltr" style="text-align: left;">
R = Conductor resistance at 65 in /m</div>
<div dir="ltr" style="text-align: left;">
I = R.M.S. value of current in each core</div>
<div dir="ltr" style="text-align: left;">
Hence the total core loss is given by n I<sup>2</sup>R.</div>
<div dir="ltr" style="text-align: left;">
The heat generated in the core of the cable passes through the dielectric medium to the sheath.</div>
<div dir="ltr" style="text-align: left;">
Let θ<sub>m</sub> = Maximum permissible of the core</div>
<div dir="ltr" style="text-align: left;">
θ<sub>s </sub>= Sheath temperature</div>
<div dir="ltr" style="text-align: left;">
Then we can write,</div>
<div dir="ltr" style="text-align: left;">
nI<sup>2</sup>R = (θ<sub>m</sub>-θ<sub>s </sub>)/S<sub>1 </sub>.................(1)</div>
<div dir="ltr" style="text-align: left;">
Where S<sub>1</sub>= Thermal resistance of the dielectric</div>
<div dir="ltr" style="text-align: left;">
Let λ = Sheath loss/ Core loss</div>
<div dir="ltr" style="text-align: left;">
Sheath loss = λ core loss</div>
<div dir="ltr" style="text-align: left;">
Total loss = Core loss + Sheath loss</div>
<div dir="ltr" style="text-align: left;">
= (1+λ) core loss = (1+λ) nI<sup>2</sup>R </div>
<div dir="ltr" style="text-align: justify;">
This is the heat flowing through bedding, serving and ground. While the
total thermal resistance of bedding, serving and ground is S<sub>4 </sub>+ S<sub>5 </sub>+G. And the corresponding temperature difference is difference between sheath temperature θ<sub>s </sub> and ambient temperature . Hence we can write,</div>
<div dir="ltr" style="text-align: left;">
(1+λ) nI<sup>2</sup>R = (θ<sub>s </sub>- θ<sub>a </sub>)/(S<sub>4 </sub>+ S<sub>5 </sub>+G) ..................(2)</div>
<div dir="ltr" style="text-align: left;">
As θ<sub>s </sub> is generally not known, eliminate θ<sub>s </sub> from (1) and (2).</div>
<div dir="ltr" style="text-align: left;">
θ<sub>m </sub>- θ<sub>s </sub>= nI<sup>2</sup>RS<sub>1</sub></div>
<div dir="ltr" style="text-align: left;">
and θ<sub>s </sub>- θ<sub>a </sub>= (1+λ) nI<sup>2</sup>R (S<sub>4 </sub>+ S<sub>5 </sub>+G)</div>
<div dir="ltr" style="text-align: left;">
θ<sub>m </sub>- θ<sub>a </sub>= nI<sup>2</sup>R <b>{</b>S<sub>1</sub>+(1+λ)(S<sub>4 </sub>+ S<sub>5 </sub>+G)<b>}</b></div>
<div class="separator" style="clear: both; text-align: center;">
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<div dir="ltr" style="text-align: left;">
<br />
</div>
<div dir="ltr" style="text-align: left;">
This is the required current carrying capacity of a cable.</div>
<div dir="ltr" style="text-align: left;">
The allowable temperature values for the various types of cables are given in the Table 1.</div>
<div class="separator" style="clear: both; text-align: center;">
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<div dir="ltr" style="text-align: left;">
<br />
</div>
<div dir="ltr" style="text-align: justify;">
Practically the
current carrying capacity also depends on the factors like grouping and
proximately of the other cables, load factor, load cycle, ambient soil
temperatures in which cable is to lay etc.</div>
<div dir="ltr" style="text-align: left;">
In practice S<sub>4</sub>, S<sub>5 </sub>and λ can be neglected hence current carrying capacity can be expressed as,</div>
<div dir="ltr" style="text-align: left;">
<br />
</div>
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Unknownnoreply@blogger.com9tag:blogger.com,1999:blog-3691402397720851976.post-82586591904979437892013-04-21T06:29:00.001-07:002013-04-21T06:30:47.795-07:00Heat Flow in a Cable As seen earlier, the current
in the cable, dielectric loss and sheath loss together cause the
increase in the temperature of the cable. The heat produced due to
increase in temperature must be dissipated to the soil. When rate of
heat generation and dissipation becomes equal then temperature becomes
constant. This temperature in fact is the important factor for deciding
current carrying capacity of the cable.<br />
<div dir="ltr" style="text-align: justify;">
The path for the heat dissipation is through the dielectric, then
sheath, bedding, serving and finally into the surrounding soil or air.
The heat flow in a 3 phase belted cable is shown in the Fig. 1(a) while
heat flow due to a three single core cable to ground is shown in the
Fig. 1(b).<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiezFcSUDJAIY0FO5E2d6uCZCehP_pOemsAv5PnwTre3Ak1Ln1wu0LDF0jw8KmayFIy0A703opxpsYdGrUBezlpq7hYRJbyxTr_2z_gODgEQvBL1JeIcgLBgvYY5s-Rp0xC8SAOc-bTEdE/s1600/ABB122.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiezFcSUDJAIY0FO5E2d6uCZCehP_pOemsAv5PnwTre3Ak1Ln1wu0LDF0jw8KmayFIy0A703opxpsYdGrUBezlpq7hYRJbyxTr_2z_gODgEQvBL1JeIcgLBgvYY5s-Rp0xC8SAOc-bTEdE/s1600/ABB122.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 1</b></td></tr>
</tbody></table>
</div>
<div dir="ltr" style="text-align: justify;">
In case of three phase belted cable, the heat flows through three
parallel paths and all the three conductors are at same temperature.</div>
<div dir="ltr" style="text-align: justify;">
The overall heat flow in a cable is similar to the flow of leakage
current in an electric circuit ; flowing radially out from core to
ground through dielectric, sheath, bedding armouring and serving.</div>
<div dir="ltr" style="color: blue; text-align: left;">
<u>1.1 Thermal Characteristics of Cable</u></div>
<div dir="ltr" style="text-align: left;">
The current in an electric circuit is given by,<br />
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Similarly heat flow H is given by,<br />
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<div style="text-align: justify;">
Thermal resistance S is defined as the resistance which allows the heat
flow of 1 watt when a temperature difference of is maintained. It is
given by,</div>
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Where k = Thermal resistivity of the material</div>
<div dir="ltr" style="text-align: left;">
<span style="font-family: Vivaldi;">l </span>= Length of the path of heat flow</div>
<div dir="ltr" style="text-align: left;">
A = Area of section through which heat flows</div>
<div dir="ltr" style="text-align: left;">
The thermal resistivity is expressed in <sup>o</sup>C/watt/cm</div>
<div dir="ltr" style="text-align: left;">
Consider the equivalent circuit for heat flow in a three phase cable as shown in the Fig. 2.<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj-z0hn2hru-yXjXEabimr1zjsRkdXkPAiOEk5dUvGjrxmFgcecADrvv7zKv25JouUPqwGgvHzjPTuPVxb4dcZWUdUBw1ewq_XanxX3GKozIY5C6-T4ccHi9uLzni2m11sZgp-NPQnxrKk/s1600/ABB126.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj-z0hn2hru-yXjXEabimr1zjsRkdXkPAiOEk5dUvGjrxmFgcecADrvv7zKv25JouUPqwGgvHzjPTuPVxb4dcZWUdUBw1ewq_XanxX3GKozIY5C6-T4ccHi9uLzni2m11sZgp-NPQnxrKk/s1600/ABB126.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 2</b></td></tr>
</tbody></table>
<div style="text-align: justify;">
The total temperature difference is difference between conductor temperature and ambient temperature. The S<sub>1</sub>, S<sub>2</sub> and S<sub>3 </sub>are
thermal resistances of the three dielectric paths which are in
parallel. Then heat flows through bedding whose thermal resistance is S<sub>4</sub>. Finally it flows through serving whose thermal resistance is S<sub>5</sub>.
The G represents the thermal resistance of ground to ambient
temperature. The current carrying capacity of cables depends on the heat
dissipation and to calculate it, it is necessary to obtain the thermal
resistances of various parts through which heat flows.</div>
</div>
<div dir="ltr" style="color: blue; text-align: left;">
<u>1.2 Thermal Resistance of Single Core Cable</u><br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEioHSwlhZILDrqdhq3DEOh5rYn_Mx0oA73u-C_JNH-A_TC7ecAJrgqqs_O2xS90lVC195QosZNK8FWy8SboXw9YUc2vN1xQFsSNb2B9zU19KwsS-FD2Vae-Tfp_nx3V-5jR8Zjbaq8Z2gY/s1600/ABB127.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEioHSwlhZILDrqdhq3DEOh5rYn_Mx0oA73u-C_JNH-A_TC7ecAJrgqqs_O2xS90lVC195QosZNK8FWy8SboXw9YUc2vN1xQFsSNb2B9zU19KwsS-FD2Vae-Tfp_nx3V-5jR8Zjbaq8Z2gY/s1600/ABB127.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 3 Single core cable</b></td></tr>
</tbody></table>
<br /></div>
<div dir="ltr" style="text-align: left;">
Consider a single core cable as shown in the Fig. 3. let</div>
<div dir="ltr" style="text-align: left;">
r = Core radius</div>
<div dir="ltr" style="text-align: left;">
R = Overall radius</div>
<div dir="ltr" style="text-align: left;">
<span style="font-family: Vivaldi;">l</span> = Length</div>
<div dir="ltr" style="text-align: left;">
k = Thermal resistivity</div>
<div dir="ltr" style="text-align: left;">
ds = (k dx)/A<br />
<div class="separator" style="clear: both; text-align: center;">
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</div>
<div dir="ltr" style="text-align: left;">
The value of k for the cables above 22 kV is 550 while it is 750 for pressure upto and including 22 kV.</div>
<div dir="ltr" style="text-align: justify;">
It is difficult to find the expression for the thermal resistance of
three core cable. The empirical formula given by Simon is,<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjmOR5f6mRXKu81n_4p_576xY44FBObCpw3NHo7bOTWfozRN4swTM5kPzKENmAxKlr-MrOcPhxonscqvQGIvDKqWzwO5PRUj8TOXolgCcPMd8yzPHb2zHGPEHaUqEEIx4zs9NxJjf28r4w/s1600/ABB130.jpeg" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjmOR5f6mRXKu81n_4p_576xY44FBObCpw3NHo7bOTWfozRN4swTM5kPzKENmAxKlr-MrOcPhxonscqvQGIvDKqWzwO5PRUj8TOXolgCcPMd8yzPHb2zHGPEHaUqEEIx4zs9NxJjf28r4w/s320/ABB130.jpeg" height="54" width="320" /></a></div>
</div>
<div dir="ltr" style="text-align: left;">
Where T = Thickness of the conductor insulation</div>
<div dir="ltr" style="text-align: left;">
t = Thickness of the belt insulation</div>
<div dir="ltr" style="text-align: left;">
r = Radius of the condcutor</div>
<div dir="ltr" style="color: blue; text-align: left;">
<u>1.3 Thermal Resistance of Soil</u></div>
<div dir="ltr" style="text-align: justify;">
The thermal resistance of the soil depends on the nature of the soil
and its moisture content. Hence exactly thermal resistance can not be
obtained for the soil. Assuming the ground to be isothermal and
homogeneous, the thermal resistance of ground is given by,<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi8JrzgqD-gY0sI_edzCGxKguvoKPi4bT_nc3DUYeaYYkBYX8XThjnTSf_-AJlhqcZSFXlyVGze8V9N0yaGCokzbyiVlZXy-YbtGeAKWvFq74OIyqTIbwSCWWPCBKyncGJQQcGKx8cUhjc/s1600/ABB131.jpeg" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi8JrzgqD-gY0sI_edzCGxKguvoKPi4bT_nc3DUYeaYYkBYX8XThjnTSf_-AJlhqcZSFXlyVGze8V9N0yaGCokzbyiVlZXy-YbtGeAKWvFq74OIyqTIbwSCWWPCBKyncGJQQcGKx8cUhjc/s1600/ABB131.jpeg" /></a></div>
</div>
<div dir="ltr" style="text-align: left;">
Where k = Thermal resistivity of soil</div>
<div dir="ltr" style="text-align: left;">
h = Depth of cable axis below the soil</div>
<div dir="ltr" style="text-align: left;">
R = Radius upto lead sheath i.e. overall radius</div>
<div style="text-align: justify;">
In practice it is necessary to multiply value of k by 2/3 which is
obtained in the laboratory, to obtain acula value of G.</div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgQp3QVihOX9p4SPB7baU4wgew0t03QRf5UysKlMu6IEQTwuPM__W96aJKcTIXnGlA6eRzppqIjkDy-Ppdmiov65RPyEwUYFS-hMaW39L8JBkoyuvKdKq-anLlU1sX7WoXu0jd60d8sHeY/s1600/ABB132.jpeg" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgQp3QVihOX9p4SPB7baU4wgew0t03QRf5UysKlMu6IEQTwuPM__W96aJKcTIXnGlA6eRzppqIjkDy-Ppdmiov65RPyEwUYFS-hMaW39L8JBkoyuvKdKq-anLlU1sX7WoXu0jd60d8sHeY/s320/ABB132.jpeg" height="42" width="320" /></a></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3691402397720851976.post-33442218379492985322013-04-21T06:28:00.001-07:002013-04-21T06:28:16.101-07:00Heating of Cables Under working conditions, the temperature of the cables increases due to the following factors,<br />
<div dir="ltr" style="text-align: left;">
1. The heat produced within the cables.</div>
<div dir="ltr" style="text-align: left;">
2. The dissipation of the heat upto the periphery of the cables.</div>
<div dir="ltr" style="text-align: left;">
3. The heat dissipation to the surrounding medium.</div>
<div dir="ltr" style="text-align: left;">
4. The current carried by the cables.</div>
<div dir="ltr" style="text-align: left;">
5. The various load conditions like continuous, distributed, intermittent etc.</div>
<div dir="ltr" style="text-align: justify;">
Out of all these factors, the heat produced within the cables is most
important from the point of view of heating of the cables. The heat is
produced within the underground cables due to following losses,</div>
<div dir="ltr" style="text-align: left;">
a. Copper loss which is also called I<sup>2</sup>R loss or core loss.</div>
<div dir="ltr" style="text-align: left;">
b. Dielectric loss.</div>
<div dir="ltr" style="text-align: left;">
c. Sheath loss.</div>
<div dir="ltr" style="color: blue; text-align: left;">
<u>1.1 Copper Loss in Cables</u></div>
<div dir="ltr" style="text-align: justify;">
The copper loss is determined by the expression I<sup>2</sup>R.
The resistance of the conductor changes as the temperature changes. The
resistance increases as the temperature increases. Hence to find copper
loss it is necessary to obtain the resistance value correctly. It is
determined by considering the following factors,</div>
<div dir="ltr" style="text-align: left;">
1. The resistance at any temperature is given by,</div>
<div class="separator" style="clear: both; text-align: center;">
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<div dir="ltr" style="text-align: left;">
Where R<sub>1</sub> = Resistance at t<sub>1</sub></div>
<div dir="ltr" style="text-align: left;">
<span dir="RTL" lang="AR-SA"> </span><span dir="RTL" lang="AR-SA"></span><sub> </sub>α<sub>1</sub> = Resistance temperature coefficient of material at t<sub>1</sub></div>
<div dir="ltr" style="text-align: left;">
Δt = Temperature rise </div>
<div dir="ltr" style="text-align: justify;">
So knowing R<sub>1</sub> i.e. resistance at ambient temperature and assuming temperature rise about of 50<sup>o</sup>C, the resistance is determined.</div>
<div dir="ltr" style="text-align: justify;">
2.
The length of outermost strand is more than the central strand. To
allow for stranding, the resistance value calculated as per as the
central strand is multiplied by 1.02.</div>
<div dir="ltr" style="text-align: justify;">
3.
The effective area of cross section is smaller than actual section
hence the resistance value is further increased by multiplying it by
1.02.</div>
<div dir="ltr" style="text-align: justify;">
Thus finally cross losses are determined as I<sup>2</sup>R where R is effective resistance considering all the factors discussed above.</div>
<div dir="ltr" style="color: blue; text-align: left;">
<u>1.2 Dielectric Loss </u></div>
<div dir="ltr" style="text-align: justify;">
There exists a capacitance between a conductor and the sheath, with a
dielectric medium in between the two. This is represented as C. The
leakage resistance is denoted as R. The equivalent circuit of the cable
is a parallel combination of R and C. So there are two currents, one
perpendicular to voltage V which is leading capacitive current I<sub>c</sub> while other is in phase which voltage V which is resistive current I<sub>d</sub> representing dielectric loss. This is shown in the Fig. 1(a) and (b).</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgv_wvo2OWokG2nU6fZHO9mXvTM4JRwWbR5lBdDd0EyAjvEF5ZfsFU2WkNlY0FMsuNcI75tJyCFjy91ZoYHufa4Rb51B1BEVZ8rlG4VMuwxqe-w8rl8QupX0158Aw8ui18jAN8_6j4m_S0/s1600/ABB118.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgv_wvo2OWokG2nU6fZHO9mXvTM4JRwWbR5lBdDd0EyAjvEF5ZfsFU2WkNlY0FMsuNcI75tJyCFjy91ZoYHufa4Rb51B1BEVZ8rlG4VMuwxqe-w8rl8QupX0158Aw8ui18jAN8_6j4m_S0/s1600/ABB118.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 1</b></td></tr>
</tbody></table>
<div dir="ltr" style="text-align: justify;">
<br />
</div>
<div dir="ltr" style="text-align: left;">
The dielectric loss is loss due to leakage resistance given by,</div>
<div dir="ltr" style="text-align: left;">
W = V<sup>2</sup>/R</div>
<div dir="ltr" style="text-align: left;">
Now tanδ = I<sub>d</sub>/I<sub>c</sub> = (V/R) / (V/X<sub>c</sub>)</div>
<div dir="ltr" style="text-align: left;">
<span dir="LTR"><b>.<sup>.</sup>.</b></span> V/R = (V/X<sub>c</sub>) tan δ = VωCtanδ</div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjH7UsQp1gliY2G8oJyI-6IjSKOHBBZRGu-Zzb2kIzKjG-gNJPtosYLVHln892HqnLK7hyphenhyphenMxYS5MJ2KBzlyYGxGPvmjdx2SrMEbnCGrZfzMMwYByHN0fz3n-Bf6WjZGjFN2zePCpUUznf0/s1600/ABB119.jpeg" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjH7UsQp1gliY2G8oJyI-6IjSKOHBBZRGu-Zzb2kIzKjG-gNJPtosYLVHln892HqnLK7hyphenhyphenMxYS5MJ2KBzlyYGxGPvmjdx2SrMEbnCGrZfzMMwYByHN0fz3n-Bf6WjZGjFN2zePCpUUznf0/s1600/ABB119.jpeg" /></a></div>
<div dir="ltr" style="text-align: left;">
<br />
</div>
<div dir="ltr" style="text-align: left;">
Where δ = Dielectric loss angle in radius</div>
<div dir="ltr" style="text-align: justify;">
Generally δ is very small and hence tan δδ. For low voltage cable
dielectric loss can be neglected as it is small but for high voltage
cables it must be considered.</div>
<div dir="ltr" style="text-align: left;">
The angle Φ<sub>d</sub> is the power factor angle of dielectric.</div>
<div dir="ltr" style="text-align: left;">
cosΦ<sub>d</sub> = cos (90-δ) = sinδ</div>
<div dir="ltr" style="text-align: left;">
It depends on the temperature and voltage stress to which dielectric is subjected.</div>
<div dir="ltr" style="color: blue; text-align: left;">
<u>1.3 Sheath Loss</u></div>
<div dir="ltr" style="text-align: justify;">
In a.c. transmission, alternating currents flowing through the cable
produce pulsating magnetic field. This electromagnetic pulsating field
links with the lead sheath and induces current in it. The value of this
current depends on the frequency of pulsating field, sheath resistance,
arrangement of cables and sheath conditions whether it is bounded or
unbounded. These sheath currents produce the sheath losses.</div>
<div dir="ltr" style="text-align: left;">
There are two types of currents in the sheath,</div>
<div dir="ltr" style="text-align: justify;">
1. Sheath eddy currents having both inword and outword directions and flow totally in the sheath of same cable.</div>
<div dir="ltr" style="text-align: left;">
2. Sheath circuit currents which flow from sheath of one cable to the sheath of other cable.</div>
<div dir="ltr" style="text-align: justify;">
The unbounded cable means having one end or no ends, electrically
shorted hence sheath circuit currents are absent in unbounded cable. The
bounded cables means the two different cables have a sheath
electrically connected at both the ends hence both the types of currents
are present in them.</div>
<div dir="ltr" style="text-align: left;">
The approximate formula to calculate sheath losses due to sheath eddy current is suggested by Arnold as,</div>
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<div dir="ltr" style="text-align: left;">
<br />
</div>
<div dir="ltr" style="text-align: left;">
Where I = Current per conductor in Amp</div>
<div dir="ltr" style="text-align: left;">
r<sub>m</sub> = Mean radius of sheath</div>
<div dir="ltr" style="text-align: left;">
R<sub>s</sub> = Sheath resistance in </div>
<div dir="ltr" style="text-align: left;">
d = Spacing between conductors</div>
<div dir="ltr" style="text-align: left;">
These losses are practically very small and hence generally neglected.</div>
<b>Key Point</b> : Thus core loss, dielectric loss and sheath loss together constitute to the heating of the cables.Unknownnoreply@blogger.com2tag:blogger.com,1999:blog-3691402397720851976.post-7382821499767498302013-04-21T06:26:00.001-07:002013-04-21T06:26:38.375-07:00Capacitance of Three Core Cables In three core cables,
capacitance play an important role because in such cables capacitances
exist between the cores as well as each core and the sheath. These
capacitances are dominating as the dielectric constant of the dielectric
used in cables is much more than the air. The capacitances are shown in
the Fig. 1.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiAHhCPLh3ZPNycEVAExLuPeJ0tUSxRhl-xaDI2QRLxIZXouUZ5nIjvUUGUR8_whiVVE1T2C5tqPTwUmQsQPDly7gjBZSmNue5jKkY7VNm78DwXi5l6STOL4WbtvztHYXFzCoBO3IlMKQs/s1600/ABB17.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiAHhCPLh3ZPNycEVAExLuPeJ0tUSxRhl-xaDI2QRLxIZXouUZ5nIjvUUGUR8_whiVVE1T2C5tqPTwUmQsQPDly7gjBZSmNue5jKkY7VNm78DwXi5l6STOL4WbtvztHYXFzCoBO3IlMKQs/s1600/ABB17.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 1</b></td></tr>
</tbody></table>
<div dir="ltr" style="text-align: left;">
The core to core capacitances are denoted as Cc while core to sheath capacitance are denoted as Cs.</div>
<div dir="ltr" style="text-align: justify;">
The core to core capacitances Cc are in delta and can be represented in the equivalent star as shown in the Fig. 2.<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiSUobL8lf2OLG0RaI7xpcPhbcLxZJrQmTjqMgCyZPx-sT4u82ktKKpBKKZOQ_n57y4wou4fOcgMLzNSL6l0aXtlbWPGo2QMjj1JhhRrLkTFxDl_rzdqEKwjRQZzEkvex6l6KhT0s9VJsQ/s1600/ABB18.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiSUobL8lf2OLG0RaI7xpcPhbcLxZJrQmTjqMgCyZPx-sT4u82ktKKpBKKZOQ_n57y4wou4fOcgMLzNSL6l0aXtlbWPGo2QMjj1JhhRrLkTFxDl_rzdqEKwjRQZzEkvex6l6KhT0s9VJsQ/s1600/ABB18.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 2</b></td></tr>
</tbody></table>
</div>
<div dir="ltr" style="text-align: left;">
The impedance between core 1 and the star point, Z<sub>1 </sub> can be obtained as,<br />
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</div>
<div dir="ltr" style="text-align: justify;">
If star point is assumed to be at earth potential and if sheath is also
earthed then the capacitance of each conductor to neutral is,<br />
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</div>
<div dir="ltr" style="text-align: left;">
If V<sub>ph </sub> is the phase voltage then charging current per phase is,<br />
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</div>
<div dir="ltr" style="color: blue; text-align: left;">
<u>1.1 Measurement of C<sub>s </sub>and C<sub>c </sub></u> </div>
<div dir="ltr" style="text-align: justify;">
The total capacitance is not easy to calculate but by actual practical measurement C<sub>s </sub> and C<sub>c </sub> can be determined.</div>
<div dir="ltr" style="text-align: left;">
Practical measurement involves two cases :</div>
<div dir="ltr" style="text-align: left;">
<b>Case 1</b> : The core 2 and 3 are connected to sheath.</div>
<div dir="ltr" style="text-align: justify;">
Thus the C<sub>c </sub> between cores 2 and 3 and C<sub>s </sub> between cores 2, 3 and sheath get eliminated as shown in the Fig. 3.<br />
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</div>
<div dir="ltr" style="text-align: left;">
All the three capacitances are now in parallel across core 1 and the sheath.</div>
<div dir="ltr" style="text-align: left;">
The capacitance of core 1 with sheath is measured practically and denoted by C<sub>a</sub>.<br />
C<sub>a </sub> = C<sub>s </sub> + 2C<sub>c </sub> ............(1)<br />
<b>Case 2 </b>: All the three cores are bundled together.</div>
<div dir="ltr" style="text-align: left;">
This eliminates all the core-core capacitances. This is shown in the Fig. 5.<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgQmpxP_kSMcblg7MR3d6s9HkzUUwvWEsd5iXL7zyctfKtDQAno06WPXY6gTUl8pu3C5wVKIJIbYLxygy0BOH-6Drg1JsErA6ynqPwJ_YBkjNLSDaGEWFg6PfFfulVFgLj4U3XuqnJuOC8/s1600/ABB114.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgQmpxP_kSMcblg7MR3d6s9HkzUUwvWEsd5iXL7zyctfKtDQAno06WPXY6gTUl8pu3C5wVKIJIbYLxygy0BOH-6Drg1JsErA6ynqPwJ_YBkjNLSDaGEWFg6PfFfulVFgLj4U3XuqnJuOC8/s1600/ABB114.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 5</b></td></tr>
</tbody></table>
The capacitances C<sub>s </sub> are in parallel between the common core and sheath.</div>
<div dir="ltr" style="text-align: left;">
This capacitance is practically measured and denoted as C<sub>b</sub>.</div>
<div dir="ltr" style="text-align: left;">
C<sub>b </sub> = 3 C<sub>s </sub> ...........(2)<br />
Solving (1) and (2) simultaneously,<br />
C<sub>a </sub> = (C<sub>b </sub>/3) + 2C<sub>c </sub> <br />
C<sub>c </sub> = (C<sub>a </sub>/2)-(C<sub>b </sub>/2) and C<sub>s </sub> = C<sub>b </sub>/3</div>
<div dir="ltr" style="text-align: left;">
Thus both the capacitances can be determined.</div>
<div dir="ltr" style="text-align: left;">
C<sub>N </sub> = C<sub>s </sub> + 3C<sub>c </sub> =(C<sub>b</sub>/3) + 3<b>(</b>(C<sub>a </sub>/2) -(C<sub>b </sub>/2)<b>)</b><br />
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<br />
<div style="color: blue;">
<u>1.2 Capacitance of Three Core Cable</u></div>
<div style="text-align: justify;">
There is one empirical formula to calculate the capacitance of a three
core belted cable, stated by Simon. It is applicable for the circular
conductors. The formula gives the capacitance of a three core cable to
neutral per phase per kilometer length of the cable. The formula is
given as,</div>
Where ε<sub>r </sub>= Relative permittivity of the dielectric<br />
d = Conductor diameter<br />
t = Belt Insulation thickness<br />
T = Conductor insulation thickness<br />
<div style="text-align: justify;">
The formula can be used when the test results are not available. This gives approximate value of the capacitance. If ε<sub>r </sub>is
not given, it can be assumed to be 3.5. It must be remembered that all
the values of d, t and T must be used in the same units while using the
formula.</div>
<div style="text-align: justify;">
<b>Example</b> : A three
core cable has core diameter 0f 2 cm and core to core distance of 4 cm.
The dielectric material has relative permittivity of 5. Compute the
capacitance of this cable per phase per km. Thickness of the conductor
insulation is 1 cm and that of belt insulation is 0.5 cm.</div>
<b>Solution </b>: d = 2 cm , ε<sub>r </sub>= 5, T = 1 cm, t = 0.5 cm<br />
Use the empirical formula as the test results are not given.<br />
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<br />
<b><a href="http://www.box.net/shared/58rtjoj0s0upoptu8lzc">Read examples on three core cables</a></b></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3691402397720851976.post-22833693409000439862013-04-21T06:24:00.001-07:002013-04-21T06:24:41.102-07:00Difficulties in Grading In the intersheath grading,
the intersheath has to be thin. Hence there is possibility of damage to
it while laying the cable. Similarly intersheath has to carry the
charging current which can cause overheating of cable.<br />
<div dir="ltr" style="text-align: justify;">
In capacitance grading, to have uniform distribution of stress it is
necessary to select the dielectrics of proper permittivities. But
practically it i difficult to get the proper values of permittivities.
But practically it is difficult to get the proper values of
permittivities. Similarly the permittivity of dielectric changes with
the time which can cause uneven distribution of stress. Such uneven
distribution may lead to breakdown at the normal operating voltage.</div>
<div dir="ltr" style="text-align: left;">
<b><a href="http://www.box.net/shared/kh6ivyr359svo6y9vtfj">Read examples on cables</a></b></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3691402397720851976.post-41153626927361539612013-04-21T06:23:00.001-07:002013-04-21T06:23:09.595-07:00Capacitance Grading The grading done by using
the layers of dielectrics having different permittivities between the
core and the sheath is called capacitance grading.<br />
<div dir="ltr" style="text-align: justify;">
In intersheath grading, the permittivity of dielectric is same
everywhere and the dielectric is said to be homogeneous. But in case of
capacitance grading, a composite dielectric is used.</div>
<div dir="ltr" style="text-align: left;">
Let d<sub>1</sub> = Diameter of the dielectric with permittivity ε<sub>1</sub></div>
<div dir="ltr" style="text-align: left;">
and D = Diameter of the dielectric with permittivity ε<sub>2</sub></div>
<div dir="ltr" style="text-align: left;">
This is shown in the Fig. 1.</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiqdlY2Uh9ypZOapADSgDxUXiB-DcpLMfk1FhVf0DKj5r5AU33OgnO2mFthMpImZNj5TzgO9iSTZ-LXw8ZoQZNA9ZGQYHtvq9qgqjApuZoQlumnNa9PV7YrkIeDKYF58HcTeFB3LyEP428/s1600/ABB16.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiqdlY2Uh9ypZOapADSgDxUXiB-DcpLMfk1FhVf0DKj5r5AU33OgnO2mFthMpImZNj5TzgO9iSTZ-LXw8ZoQZNA9ZGQYHtvq9qgqjApuZoQlumnNa9PV7YrkIeDKYF58HcTeFB3LyEP428/s1600/ABB16.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 1 Capacitance grading</b></td></tr>
</tbody></table>
<div dir="ltr" style="text-align: left;">
<br />
</div>
<div dir="ltr" style="text-align: left;">
The stress at a point which is at a distance x is inversely proportional to the distance x and given by,</div>
<div dir="ltr" style="text-align: left;">
g<sub>x </sub>= Q/(2πε x)</div>
<div dir="ltr" style="text-align: left;">
Hence the stress at point in the inner dielectric is,</div>
<div dir="ltr" style="text-align: left;">
g<sub>1 </sub>= Q/(2πε<sub>1</sub> x)</div>
<div dir="ltr" style="text-align: left;">
Similarly the dielectric stress in the outer dielectric is,</div>
<div dir="ltr" style="text-align: left;">
g<sub>2 </sub>= Q/(2πε<sub>2</sub> x)</div>
<div dir="ltr" style="text-align: left;">
Hence the total voltage V can be expressed as,</div>
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<div dir="ltr" style="text-align: left;">
The stress is maximum at surface of conductor i.e. x =d/2.</div>
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<div dir="ltr" style="text-align: left;">
And the stress is maximum at inner surface of dielectric i.e. at x = d<sub>1</sub>/2.</div>
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<div dir="ltr" style="text-align: left;">
Substituting Q interms of V we get,</div>
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<div dir="ltr" style="text-align: left;">
<br />
</div>
<div dir="ltr" style="text-align: justify;">
<b>Key Point </b>: Thus the electric stress is inversely proportional to the permittivities and the inner radii of the dielectrics.</div>
<div dir="ltr" style="color: blue; text-align: left;">
<u>1.1 Condition for Equal Maximum Stress</u></div>
<div dir="ltr" style="text-align: left;">
Let us obtain the condition under which the maximum values of the stresses in the two regions are equal.</div>
<div dir="ltr" style="text-align: left;">
The maximum stresses are given by,</div>
<div dir="ltr" style="text-align: left;">
g<sub>1max </sub>= Q/(πε<sub>1</sub>d)</div>
<div dir="ltr" style="text-align: left;">
and g<sub>2max</sub> = Q/(πε<sub>2</sub>d<sub>1</sub>)</div>
<div dir="ltr" style="text-align: left;">
Equating the two stresses,</div>
<div dir="ltr" style="text-align: left;">
Q/(πε<sub>1</sub>d) = Q/(πε<sub>2</sub>d<sub>1</sub>)</div>
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<div dir="ltr" style="text-align: left;">
<br />
</div>
<div dir="ltr" style="text-align: left;">
Now d<sub>1</sub> is greater than d so to satisfy above equation ε<sub>2</sub> must be less than ε<sub>1</sub>.</div>
<div dir="ltr" style="text-align: left;">
Thus the dielectric nearest to the conductor must have the highest permittivity.</div>
<div dir="ltr" style="text-align: justify;">
Similar for the grading with three dielectrics with permittivities ε<sub>1</sub>, ε<sub>2 </sub>and ε<sub>3</sub>, for equal maximum stress the condition is,</div>
<div dir="ltr" style="text-align: left;">
And </div>
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<div dir="ltr" style="text-align: left;">
<br />
</div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3691402397720851976.post-80799893736934497122013-04-21T06:20:00.001-07:002013-04-21T06:20:23.796-07:00Use of Intersheath for Grading In this method of grading,
in between the core and the lead sheath number of metallic sheaths are
placed which are called intersheaths. All these intersheaths are
maintained at different potentials by connecting them to the tappings of
the transformer secondary. These potentials are between the core
potential and earth potential. Generally lead is used for these sheaths
as it is flexible and corrosion resistance but as its mechanical
strength is less, aluminium also can be used. Aluminium is low weight
and mechanically strong but it is much costlier than lead.<br />
<div dir="ltr" style="text-align: justify;">
Using the intersheaths, maintaining at different potential, uniform distribution of stress is obtained in the cables.</div>
<div dir="ltr" style="text-align: justify;">
Consider a cable with core diameter d and overall diameter with lead
sheath as D. Let two intersheaths are used having diameter d<sub>1</sub> and d<sub>2</sub> which are kept at the potentials V<sub>1</sub> and V<sub>2</sub> respectively.</div>
<div dir="ltr" style="text-align: left;">
The intersheaths and stress distribution is shown in the Fig. 1.</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEggdxblBc5pNgx6OKEymC_kTOVXZLilLD-RCIBkjAuJdEooBncQIr0bCDxbDBuZ3fODNrA6Zzo34HRdswQAuXd2MIG0BZh-1FEOb9MkgFYrdVTVZ4dU0MpQ3AtzKVhB5XraqVnNkoqXkj8/s1600/ABB125.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEggdxblBc5pNgx6OKEymC_kTOVXZLilLD-RCIBkjAuJdEooBncQIr0bCDxbDBuZ3fODNrA6Zzo34HRdswQAuXd2MIG0BZh-1FEOb9MkgFYrdVTVZ4dU0MpQ3AtzKVhB5XraqVnNkoqXkj8/s1600/ABB125.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 1</b></td></tr>
</tbody></table>
<div dir="ltr" style="text-align: left;">
<br />
</div>
<div dir="ltr" style="text-align: left;">
Let V<sub>1</sub> = Voltage of intersheath 1 with respect to earth</div>
<div dir="ltr" style="text-align: left;">
V<sub>2</sub> = Voltage of intersheath 2 with respect to earth</div>
<div dir="ltr" style="text-align: justify;">
It has been proved that stress at a point which is at a distance x is inversely proportional to distance x and given by,</div>
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<div dir="ltr" style="text-align: justify;">
<br />
</div>
<div dir="ltr" style="text-align: left;">
Where k is constant.</div>
<div dir="ltr" style="text-align: left;">
So electric stress between he conductor and intersheath 1 is,</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhcqEFduKrPSygvTIeTdyOJ9PtV_zA0abNjCcO2v0OG8AoiAQkJFgfP13vbwED3deH_g0vSF3HD4-D95xK4TbeWoqAMjxl6ndEHx-q8BwFww8iBPvK_KotRKKBY8A2LJ_vpsyv8cdxK_gM/s1600/ABB127.jpeg" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhcqEFduKrPSygvTIeTdyOJ9PtV_zA0abNjCcO2v0OG8AoiAQkJFgfP13vbwED3deH_g0vSF3HD4-D95xK4TbeWoqAMjxl6ndEHx-q8BwFww8iBPvK_KotRKKBY8A2LJ_vpsyv8cdxK_gM/s1600/ABB127.jpeg" /></a></div>
<div dir="ltr" style="text-align: left;">
Now potential difference between core and the first intersheath is V-V<sub>1</sub>.</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhQ1LFRT0d-LqWA6x9DZ_oOjgkMdHDkzg8qebEsdU4gM75eMpsP0cgpOkpfHiJp1_8dG9JR5wlnktNW3CaTuXPXyfHezpY0LWzvhZhf1SuLMY4GMTPm9wshaF9c4sTjs_QlcvsnfOl74fc/s1600/ABB128.jpeg" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhQ1LFRT0d-LqWA6x9DZ_oOjgkMdHDkzg8qebEsdU4gM75eMpsP0cgpOkpfHiJp1_8dG9JR5wlnktNW3CaTuXPXyfHezpY0LWzvhZhf1SuLMY4GMTPm9wshaF9c4sTjs_QlcvsnfOl74fc/s1600/ABB128.jpeg" /></a></div>
<div dir="ltr" style="text-align: left;">
Substituting in equation (2) we get,</div>
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<div dir="ltr" style="text-align: left;">
Now this stress is maximum at x = d/2, on core surface.</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiAlw3Rgo4j7hZNi_nbWWzzK9NXOQVpxqHHhXxGTFYrp7Z6se6DoSd2a5agq4VgEBdOMrJ6x_XPSGtG_bAgF03yZz5Dc9syq1_aSQxKwp1uEMN510CNXwKl_2sBCnS5fLszP7KCw6g23TY/s1600/ABB130.jpeg" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiAlw3Rgo4j7hZNi_nbWWzzK9NXOQVpxqHHhXxGTFYrp7Z6se6DoSd2a5agq4VgEBdOMrJ6x_XPSGtG_bAgF03yZz5Dc9syq1_aSQxKwp1uEMN510CNXwKl_2sBCnS5fLszP7KCw6g23TY/s1600/ABB130.jpeg" /></a></div>
<div dir="ltr" style="text-align: left;">
Similarly potential difference between intersheath 1 and intersheath 2 is V<sub>1</sub> - V<sub>2</sub>.</div>
<div class="separator" style="clear: both; text-align: center;">
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<div dir="ltr" style="text-align: left;">
Now g<sub>1 </sub>will be maximum at the surface of intersheath 1 i.e. x = d<sub>1</sub>/2.</div>
<div class="separator" style="clear: both; text-align: center;">
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<div dir="ltr" style="text-align: justify;">
The potential difference between intersheath 2 and outermost sheath V<sub>2</sub> is only as potential of intersheath is maintained at V<sub>2 </sub>with respect to earth.</div>
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<div dir="ltr" style="text-align: left;">
This g<sub>3 </sub>will be maximum at x = d<sub>2</sub>/2</div>
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<div dir="ltr" style="text-align: left;">
Choosing proper values of V<sub>1 </sub>and V<sub>2</sub>, g<sub>1max</sub>, g<sub>2max</sub> etc. can be made equal and hence uniform distribution of stress can be obtained.</div>
<div dir="ltr" style="text-align: justify;">
The stress can be made to vary between same maximum and minimum values as shown in the Fig. 1, by choosing d<sub>1</sub> and d<sub>2</sub> such that,</div>
<div dir="ltr" style="text-align: justify;">
d<sub>1</sub>/d = d<sub>2</sub>/d<sub>1</sub> = D/d<sub>2</sub> = <span dir="RTL" lang="AR-SA">α</span></div>
<div dir="ltr" style="text-align: justify;">
and g<sub>1max </sub>= g<sub>2max</sub> = g<sub>3max</sub></div>
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<div dir="ltr" style="text-align: left;">
Let us try the express voltages V<sub>1 </sub>and V<sub>2 </sub> interms of V and <span dir="RTL" lang="AR-SA">α</span>.</div>
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<div dir="ltr" style="text-align: left;">
Substituting value of V<sub>2 </sub> from equation (11),</div>
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</div>
<b><a href="http://www.box.net/shared/ifzag322mxokm6ole0ea">Read examples on cables</a></b>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3691402397720851976.post-72891529993386821982013-04-21T06:18:00.001-07:002013-04-21T06:18:43.580-07:00Grading of Cables We have seen that thew
stress in the insulation is maximum at the conductor surface and minimum
at the sheath. To avoid the breakdown of the insulation, it is
necessary to have uniform distribution of stress all along the
insulation.<br />
<div dir="ltr" style="text-align: justify;">
Practically some methods are used to obtain uniform distribution of
stress. The process of obtaining uniform distribution of stress in the
insulation of cables is called grading of cables.</div>
<div dir="ltr" style="text-align: left;">
The unequal distribution of stress has two effects,</div>
<div dir="ltr" style="text-align: left;">
1. Greater insulation thickness is required, which increases the cost and size.</div>
<div dir="ltr" style="text-align: left;">
2. It may lead to the breakdown of insulation.</div>
<div dir="ltr" style="text-align: left;">
Hence the grading of cables is done.</div>
<div dir="ltr" style="text-align: left;">
There are two methods of grading the cables which are,</div>
<div dir="ltr" style="text-align: left;">
1. Use of intersheaths for grading </div>
<div dir="ltr" style="text-align: left;">
2. Capacitance grading</div>
<div dir="ltr" style="text-align: left;">
Let us discuss these two grading methods in detail<br />
<br />
<u>Related Posts :</u><br />
<b><a href="http://yourelectrichome.blogspot.com/2011/09/use-of-intersheath-for-grading.html">Use of intersheath for Grading</a></b><br />
<b><a href="http://yourelectrichome.blogspot.com/2011/09/capacitance-grading.html">Capacitance Grading</a></b><br />
<b><a href="http://yourelectrichome.blogspot.com/2011/09/difficulties-in-grading.html">Difficulties in Grading</a></b></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3691402397720851976.post-90571857576852009352013-04-21T06:17:00.001-07:002013-04-21T06:17:02.268-07:00Capacitance of Single Core Cable A single core cable is
equivalent to two long co-axial cylinders. The inner cylinder is the
conductor itself while the outer cylinder is the lead sheath. The lead
sheath is always at earth potential.<br />
<div dir="ltr" style="text-align: left;">
Let d = Conductor diameter</div>
<div dir="ltr" style="text-align: left;">
D = Total diameter with sheath</div>
<div dir="ltr" style="text-align: left;">
The co-axial cylindrical from of cable and its section are shown in the Fig. 1(a) and (b).<br />
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgN-FMRYGvc8510crInIkt4r-dBuqPI534IqokpOjV6EDpUl0hWysabM3RMZH5v-D5C89mf57xYAYpt5CC9a9yx9S8AkeAGgSfg4MOJkOb6R1GxkEkwsHqNcMMOI6iBbOblnhIcq3dnKio/s1600/ABB18.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgN-FMRYGvc8510crInIkt4r-dBuqPI534IqokpOjV6EDpUl0hWysabM3RMZH5v-D5C89mf57xYAYpt5CC9a9yx9S8AkeAGgSfg4MOJkOb6R1GxkEkwsHqNcMMOI6iBbOblnhIcq3dnKio/s1600/ABB18.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 1 Capacitance of single core cable</b></td></tr>
</tbody></table>
</div>
<div dir="ltr" style="text-align: left;">
Let Q = Charge per meter length of conductor in coulombs</div>
<div dir="ltr" style="text-align: left;">
ε = Permittivity of material between core and sheath</div>
<div dir="ltr" style="text-align: left;">
Now ε = ε<sub>o</sub> ε<sub>r</sub></div>
<div dir="ltr" style="text-align: left;">
Where ε<sub>o</sub> = permittivity of free space = 8.854 x 10<sup>-12</sup> F/m</div>
<div dir="ltr" style="text-align: left;">
and ε<sub>r </sub>= Relative permittivity of the medium</div>
<div dir="ltr" style="text-align: left;">
Consider an elementary cylinder with radius x and axial length of 1 m. The thickness of the cylinder is dx.</div>
<div dir="ltr" style="text-align: justify;">
According to Gauss's theorem, the lines of flux emanating due to charge
Q on the conductor are in parallel direction and total flux line are
equal to the total charge possessed i.e. Q lines. As lines are in radial
direction, the cross-sectional area through which lines pass is surface
area. For a cylinder with radius x, the surface area is (2πx x axial
length) m<sup>2</sup>. As axial length considered is 1 m, the surface area is 2 π x m<sup>2</sup>.<br />
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The electric field intensity at any point P on the elementary cylinder is given by,</div>
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<div dir="ltr" style="text-align: left;">
g<sub>x </sub>= Dx/ε where Dx = Electric flux density <br />
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<div style="text-align: justify;">
Hence the work done in moving a unit charge through a distance dx in the direction of an electric field is g<sub>x </sub>dx.</div>
<div style="text-align: justify;">
Therefore the work done in moving a unit charge from the conductor to
sheath is the potential difference between the conductor and the sheath
given by,</div>
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The capacitance of a cable is given by,<br />
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Note that as length considered is 1 m, the capacitance is F/m.<br />
If required for length '<span style="font-family: Vivaldi;">l</span>' multiply c by '<span style="font-family: Vivaldi;">l</span>'.<br />
Substituting value of ε<sub>o</sub>,<br />
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If the length <span style="font-family: Vivaldi;">l </span>of cable is known then the total capacitance of cable is,<br />
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<div style="text-align: justify;">
<b>Note </b>:
To avoid the confusion of units, students can use the expression given
by equation (1), to calculate capacitance while solving the problem.</div>
<b>Charging current </b>: When the capacitance C of a cable is known then its reactance is given by,<br />
X<sub>c </sub>= 1/(ωC) = 1/(2πfC) Ω<br />
Then the charging current of the cable is given by,<br />
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Where V<sub>ph </sub>= Phase voltage between core and sheath = V<sub>line</sub>/√3<br />
<div style="color: blue;">
<u>1.1 Stress in Insulation</u></div>
The electrical stress in insulation is the electric field intensity acting at any point P in insulation.<br />
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The stress is maximum at the surface of the conductor i.e. when x = r.<br />
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Similar the minimum stress will be at the length i.e. x = R hence<br />
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The variation of stress in dielectric material is shown ion the Fig. 2.<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiN3_XrnAuNzCK_u7Cl40tbSH4RHos1NYCMJcehid7seFrdxplzKGV9KdkUGVhiaxJRyKZFG4HmFbKU94k4LExihMKAPojA196yKDzhkEnqOPE8d9Sq-9T9unOVGstvi7GZWaPV3WJuZ48/s1600/ABB144.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiN3_XrnAuNzCK_u7Cl40tbSH4RHos1NYCMJcehid7seFrdxplzKGV9KdkUGVhiaxJRyKZFG4HmFbKU94k4LExihMKAPojA196yKDzhkEnqOPE8d9Sq-9T9unOVGstvi7GZWaPV3WJuZ48/s1600/ABB144.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 2</b></td></tr>
</tbody></table>
<br />
The ratio of maximum and minimum stress is,<br />
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<b>Key Point </b>:
If value of voltage used is r.m.s. we get r.m.s. values of stresses and
if value of voltage used is peak, we get peak values of stresses.<br />
<div style="color: blue;">
<u>1.2 Economical core Diameter</u></div>
In
practice, the maximum stress value should be as low as possible. When
the voltage V and sheath diameter D are fixed, the only parameter to be
selected is the core diameter d. So d should be selected for which value
is minimum.<br />
The value of will be minimum when ∂g<sub>max</sub>/∂d = 0<br />
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Now the value of ∂g<sub>max</sub>/∂d must be zero to get minimum g<sub>max</sub>.<br />
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<b>Key Point </b>: The core diameter must be 1/2.718 times the sheath diameter D so as to give the minimum value of g<sub>max</sub>.<br />
The value of minimum g<sub>max</sub> is,<br />
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<span dir="LTR"><b>.<sup>.</sup>.</b></span> Minimum g<sub>max </sub>= V/r ..... As r = d/2<br />
<div style="text-align: justify;">
For high voltage cable, for a
required if d is determined by the expression (5), it gives very large
values of d than required for current carrying capacity. And such extra
copper required can increase the cost tremendously. Hence ti increase d
without the use of an extra copper following methods are used :</div>
<div style="text-align: justify;">
1. Aluminium is used instead of copper as the aluminium size is more than copper for the same current carrying capacity.</div>
2. Using stranded copper conductors around a dummy core of tube instead of hemp.<br />
3. Using stranded copper conductors around a lead tube instead of hemp.<br />
<b><a href="http://www.box.net/shared/ur7xxtzkv2obmmqht4jc">Read examples on cables </a></b></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3691402397720851976.post-22484110366473910892013-04-21T06:14:00.001-07:002013-04-21T06:14:54.572-07:00Outdoor Dead Tank Circuit Breaker - VOX <div style="text-align: justify;">
The
VOX dead tank circuit breaker is a cost-effective and reliable
solution for substations. It performs well in harsh ambient conditions
and requires limited maintenance.</div>
<div class="richText">
<ul dir="ltr" style="text-align: left;">
<li>Ratings up to 38 kV / 2000 A / 40 kA.</li>
<li>Rated impulse withstand voltage up to 200 kV peak full wave and 258 kV peak 2µs chopped wave.</li>
<li>Rated for -40°C to +50°C (standard), -65°C to +55°C (option).</li>
<li>No de-rating required for altitudes up to 10000 ft (3000m).</li>
<li>Internal arc fault and seismic endurance tested.</li>
<li>EPDM rubber bushings.</li>
<li>Large CT housing</li>
<li>Surge Arresters, VT or recloser configuration in option</li>
<li>Meets IEC, ANSI/IEEE, BS, GOST, AS, GB regulations.</li>
</ul>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhiLWCLQfwxA-VPh7oDrg9-0ELznWFUd2-FZbwTn4Gu3CZqSZjM0ouKjcQ7QadeJ7A19FDxcHCSYh3iuU_vhfMhq7Id1p4UkYIzZlbq52e5TMGDyNOr-LDPIyMLeWatM7BueUVjCb25HMWX/s1600/VOX-main.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhiLWCLQfwxA-VPh7oDrg9-0ELznWFUd2-FZbwTn4Gu3CZqSZjM0ouKjcQ7QadeJ7A19FDxcHCSYh3iuU_vhfMhq7Id1p4UkYIzZlbq52e5TMGDyNOr-LDPIyMLeWatM7BueUVjCb25HMWX/s1600/VOX-main.jpg" /></a></div>
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</div>
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</div>
<div dir="ltr" style="text-align: left;">
<b style="color: blue;">Downloads </b>: </div>
<div dir="ltr" style="text-align: left;">
-Tech publications</div>
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</a></div>
<div dir="ltr" style="text-align: justify;">
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</div>
<div dir="ltr" style="text-align: justify;">
<b>Reference </b>: <a href="http://www2.schneider-electric.com/sites/corporate/en/products-services/energy-distribution/products-offer/range-presentation.page?c_filepath=/templatedata/Offer_Presentation/3_Range_Datasheet/data/en/shared/energy_distribution/vox.xml&p_function_id=20022&p_family_id=30115&p_range_id=60673&f=NNM1%3AMV+Circuit+Breakers%7E%21NNM2%3AOutdoor+Circuit+Breakers%7E%21NNM3:VOX">Schneider Electric</a> (Products and Services > MV Circuit Breakers > Indoor Circuit Breakers > Outdoor Dead Tank Circuit Breaker - VOX)<br />
<br />
<b style="color: blue;">Downloads </b><br />
<b><a href="http://www.box.com/s/vtrk3ze1z3td57oqn2k7" target="_blank">Outdoor Dead Tank Circuit Breaker - VOX</a></b></div>
</div>
<ins style="border: none; display: inline-table; height: 280px; margin: 0; padding: 0; position: relative; visibility: visible; width: 336px;"><ins id="aswift_1_anchor" style="border: none; display: block; height: 280px; margin: 0; padding: 0; position: relative; visibility: visible; width: 336px;"></ins></ins>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3691402397720851976.post-61510962495905828022013-04-20T23:19:00.001-07:002013-04-20T23:19:26.599-07:00Insulation Resistance of a Cable The Fig 1 shows the section of a single core cable which is insulated with the help of layer of an insulating material.<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiqHIBwTLByVJaynEIhLdbRa6ow0jj9iwujc1F-Piiezh-eiHChIVZW-ZQFShvvYwB5bUgcWQWyiLXszkQcSZRzJ3RTDEbRlJD72FvLPbDmOE8xpP_OYUy687t9t_rkbh48nAsi7rgzwP4/s1600/ABB1.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiqHIBwTLByVJaynEIhLdbRa6ow0jj9iwujc1F-Piiezh-eiHChIVZW-ZQFShvvYwB5bUgcWQWyiLXszkQcSZRzJ3RTDEbRlJD72FvLPbDmOE8xpP_OYUy687t9t_rkbh48nAsi7rgzwP4/s1600/ABB1.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 1 Single core cable</b></td></tr>
</tbody></table>
<div dir="ltr" style="text-align: justify;">
<br />
</div>
<div dir="ltr" style="text-align: justify;">
In such cables,
the leakage current flows radially from centre towards the surface as
shown in the Fig.1. Hence the cross-section of the path of such current
is not constant but changes with its length. The resistance offered by
cable to path of the leakage current is called an insulation resistance
consider an elementary section of the cylindrical cable of radius x and
the thickness dx as shown in the Fig. 2. Let us find the resistance of
this elementary ring.</div>
<div dir="ltr" style="text-align: left;">
Let d = Diameter of core</div>
<div dir="ltr" style="text-align: left;">
r = d/2 = Radius of core</div>
<div dir="ltr" style="text-align: left;">
D = Diameter with sheath</div>
<div dir="ltr" style="text-align: left;">
R = D/2 = Radius of cable with sheath</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjFw562sa2npSqR8lEQzgnwcBJiltwd0YvVFHC1Dy7KpJ_ySZoj1xxP0r1b_kRwUsYZvCOeAXFZsUzRIlDjKbV0YuJeyBfJH-qaqLMtxksTGyGGV6HEqmPzv3plPQ470BrHJu-45sIb2h4/s1600/ABB11.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjFw562sa2npSqR8lEQzgnwcBJiltwd0YvVFHC1Dy7KpJ_ySZoj1xxP0r1b_kRwUsYZvCOeAXFZsUzRIlDjKbV0YuJeyBfJH-qaqLMtxksTGyGGV6HEqmPzv3plPQ470BrHJu-45sIb2h4/s1600/ABB11.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 2 Elementary ring</b></td></tr>
</tbody></table>
<div dir="ltr" style="text-align: left;">
<br />
</div>
<div dir="ltr" style="text-align: justify;">
As the leakage
current flows radially outwards, the length along which the current
flows in an elementary ring is dx. While the cross-sectional area
perpendicular to the flow of current depends on the length of <span style="font-family: Vivaldi;">l</span> of the cable.</div>
<div dir="ltr" style="text-align: left;">
Cross-section area = Surface area for length <span style="font-family: Vivaldi;">l</span> of cable </div>
<div dir="ltr" style="text-align: left;">
= (2 π x) x <span style="font-family: Vivaldi;">l</span></div>
<div dir="ltr" style="text-align: left;">
Hence the resistance of this elementary cylindrical shell is,</div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiMGv2d8BFhyphenhyphenAg3nA55yVNzA4H8YFeyWGT-8hgfULFPHDh-vLPAgeCW1eFT38W1BvR44G7D1lEUB7yompVAjpycuIPS3EJohJAes-1yb-hVi3i31OzkFDKlWiC7rEUK-Zj8FGPslTMuICg/s1600/ABB12.jpeg" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiMGv2d8BFhyphenhyphenAg3nA55yVNzA4H8YFeyWGT-8hgfULFPHDh-vLPAgeCW1eFT38W1BvR44G7D1lEUB7yompVAjpycuIPS3EJohJAes-1yb-hVi3i31OzkFDKlWiC7rEUK-Zj8FGPslTMuICg/s1600/ABB12.jpeg" /></a></div>
<div dir="ltr" style="text-align: left;">
<br />
</div>
<div dir="ltr" style="text-align: left;">
where ρ = Resistivity of the insulating material</div>
<div dir="ltr" style="text-align: justify;">
The total insulation resistance of the cable can be obtained by
integrating the resistance of an elementary ring from inner radius upto
the outer radius i.e. r to R.</div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjmeCPSM0DiORX-6ZYOs1rBFHGZs_einVxkEL3fNoSEtdHzPgCjpG_4mgdyEzTXt9T7HrPNQXOovnv1OOsEPQXTZ42aS7Wi_iPs2n4KCH3hjxCgV5p_SVv7axQIJhwFUjfRGojaS7oG5bg/s1600/ABB13.jpeg" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjmeCPSM0DiORX-6ZYOs1rBFHGZs_einVxkEL3fNoSEtdHzPgCjpG_4mgdyEzTXt9T7HrPNQXOovnv1OOsEPQXTZ42aS7Wi_iPs2n4KCH3hjxCgV5p_SVv7axQIJhwFUjfRGojaS7oG5bg/s1600/ABB13.jpeg" /></a></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjfxJdCCCz8GZS8u9BS9tE18ITxQkUB_WkhU-WwCHCQyDsmMffNYapJ5bGk7ZPG_g7WatwiooF0CLM5sSgOO1BgNiLRXASnkez6kR_rffcimjUOYRjTHI_Z5BNlaTMvtv2cluublN0754g/s1600/ABB15.jpeg" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjfxJdCCCz8GZS8u9BS9tE18ITxQkUB_WkhU-WwCHCQyDsmMffNYapJ5bGk7ZPG_g7WatwiooF0CLM5sSgOO1BgNiLRXASnkez6kR_rffcimjUOYRjTHI_Z5BNlaTMvtv2cluublN0754g/s1600/ABB15.jpeg" /></a></div>
<div dir="ltr" style="text-align: justify;">
<br />
</div>
<div dir="ltr" style="text-align: left;">
This can be expressed interms of diameters as,</div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiqBYsDG7ngN29PhmArMKo3zNlUIS42-GqO5Pq-64uT7QEh7axjV_a33dlB1GBRkzSdAy63zOVVdmrJ_Hpqn-QyiDSP-lpKweN41COVrFBVUpi8viQ_iJXU3dXuUIrvE6VfmEREQMMk04k/s1600/ABB17.jpeg" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiqBYsDG7ngN29PhmArMKo3zNlUIS42-GqO5Pq-64uT7QEh7axjV_a33dlB1GBRkzSdAy63zOVVdmrJ_Hpqn-QyiDSP-lpKweN41COVrFBVUpi8viQ_iJXU3dXuUIrvE6VfmEREQMMk04k/s1600/ABB17.jpeg" /></a></div>
<div dir="ltr" style="text-align: left;">
<br />
</div>
<div dir="ltr" style="text-align: justify;">
The value of R<sub>i </sub>
is always very high. The expression shows that the insulation
resistance is inversely proportional to its length. So as the cable
length increases, the insulation resistance decreases.</div>
<div dir="ltr" style="text-align: justify;">
This shows that if two cables are joined in series then total length
increases and hence their conductor resistances are in series giving
higher resistance but insulation resistance are in parallel decreasing
the effective insulation resistance. Thus if two cables are connected in
parallel, conductor resistances get connected in parallel while the
insulation resistance get connected in series.<br />
<b><a href="http://www.box.net/shared/ej1bsg8dz5n0cvix6e95">Read this example </a></b></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3691402397720851976.post-63542205701100286332013-04-20T23:15:00.001-07:002013-04-20T23:15:36.606-07:00Insulating Materials for Cables Number
of layers of the various materials is used around the actual conductor
in a cable. To isolate the conductor from the surroundings, the
conductor is provided with an insulation around it. The materials like
paper, vulcanized rubber, PVC etc. are used for providing such an
insulation.<br />
<div dir="ltr" style="text-align: left;">
The material to be used as an insulation must have the following properties,</div>
<div dir="ltr" style="text-align: left;">
1. To prevent leakage current, its insulation resistance must be very high.</div>
<div dir="ltr" style="text-align: left;">
2. To avoid electrical breakdown, its dielectric strength must very high.</div>
<div dir="ltr" style="text-align: left;">
3. To withstand the mechanical injures, it must be mechanically very strong.</div>
<div dir="ltr" style="text-align: left;">
4. It should be flexible.</div>
<div dir="ltr" style="text-align: left;">
5. It should be non-hygroscopic so that it will not absorb the moisture from the surroundings.</div>
<div dir="ltr" style="text-align: left;">
6. It should be non-inflammable.</div>
<div dir="ltr" style="text-align: left;">
7. It should be unaffected by acids and alkalies.</div>
<div dir="ltr" style="text-align: left;">
8. It should be capable of withstanding high breakdown voltages.</div>
<div dir="ltr" style="text-align: left;">
9. It should have high temperature withstanding capability.</div>
<div dir="ltr" style="text-align: justify;">
Practically it is not possible to have all these properties in a single
material. Hence insulation material is selected depending upon the use
of the cable and the quality of insulation required for it. Some changes
are done at the time of design depending upon the nature of material
selected. For example if the material is hygroscopic then a layer of a
waterproof covering is provided around it so that moisture can not reach
the insulation. The main insulating materials which are in used are,</div>
<div dir="ltr" style="text-align: left;">
1. Poly Vinyl Chloride (PVC)</div>
<div dir="ltr" style="text-align: left;">
2. Paper</div>
<div dir="ltr" style="text-align: left;">
3. Cross Linked Polythene</div>
<div dir="ltr" style="text-align: left;">
4. Vulcanized India Rubber (VIR)<br />
<u style="color: blue;">1.1 Poly Vinyl Chloride (PVC) </u><br />
<div dir="ltr" style="text-align: justify;">
It is thermo plastic
synthetic compound. It is available in the powder from and is obtained
from polymerisation of acetylene. This powder is chemically inert,
non-inflammable, odourless, tasteless and insoluble. It is combined with
plastic compound and a gel is used over the conductor to obtain the
insulation.</div>
<div dir="ltr" style="text-align: left;">
It has following characteristics,</div>
<div dir="ltr" style="text-align: left;">
1. Good dielectric strength of 17 kV/mm</div>
<div dir="ltr" style="text-align: left;">
2. Chemically inert.</div>
<div dir="ltr" style="text-align: left;">
3. Non-hygroscopic.</div>
<div dir="ltr" style="text-align: left;">
4. Resistant to corrosion.</div>
<div dir="ltr" style="text-align: left;">
5. Maximum continuous temperature rating of 75<sup>o</sup>C.</div>
<div dir="ltr" style="text-align: left;">
6. High electrical resistivity.</div>
<div dir="ltr" style="text-align: justify;">
The mechanical properties like elasticity of PVC are not as good as
rubber so PVC cables are used for low and medium voltage domestic,
industrial lights and power installations.</div>
<div dir="ltr" style="color: blue; text-align: left;">
<u>1.2 Paper </u></div>
<div dir="ltr" style="text-align: justify;">
The paper is very cheap insulating material. Its dielectric strength
is also high but it is hygroscopic in nature. When it is dry its
insulation resistance is very high but a small amount of moisture
reduces its insulation resistance to a very low value. Thus it is
impregnated in an insulation oil. After impregnated also it has a
tendency to absorb the moisture. Hence paper cables are never left
unsealed and provided with the protective covering. When not in use,
paper cable ends are temporarily covered with wax or tar.</div>
<div dir="ltr" style="text-align: left;">
The paper has following characteristics,</div>
<div dir="ltr" style="text-align: left;">
1. High dielectric strength of 20 kV/mm.</div>
<div dir="ltr" style="text-align: left;">
2. Higher thermal conductivity.</div>
<div dir="ltr" style="text-align: left;">
3. Low capacitance</div>
<div dir="ltr" style="text-align: left;">
4. High durability</div>
<div dir="ltr" style="text-align: left;">
5. Low cost</div>
<div dir="ltr" style="text-align: left;">
6. Maximum continuous temperature rating of 80<sup>o</sup>C.</div>
<div dir="ltr" style="text-align: left;">
7. High insulation resistance when dry.</div>
<div dir="ltr" style="text-align: justify;">
It is used in high voltage power cable manufacturing. The paper cables
are preferred when the cable route has very few joints and hence
generally used for low voltage distribution in thickly populated areas.</div>
<div dir="ltr" style="color: blue; text-align: left;">
<u>1.3 Cross Linked Polythelene</u></div>
<div dir="ltr" style="text-align: left;">
The cable using cross linked polythelene as the insulating material are called XLPE cables.</div>
<div dir="ltr" style="text-align: justify;">
The low density polythelene is treated specially due to which there
occurs cross linking of carbon atoms in it. This results into a new
material which has following properties,</div>
<div dir="ltr" style="text-align: left;">
1. High dielectric strength of 20 to 40 kV/mm</div>
<div dir="ltr" style="text-align: left;">
2. Non-inflammable : If at all the continuous flame is applied its burning stops after very few centimeters away from the flame.</div>
<div dir="ltr" style="text-align: left;">
3. Extremely high melting point.</div>
<div dir="ltr" style="text-align: left;">
4. Light in weight and flexible.</div>
<div dir="ltr" style="text-align: left;">
5. Mechanically strong.</div>
<div dir="ltr" style="text-align: left;">
6. High temperature withstanding capability.</div>
<div dir="ltr" style="text-align: left;">
7. Low moisture absorption.</div>
<div dir="ltr" style="text-align: left;">
8. Maximum continuous temperature rating of 90<sup>o</sup>C.</div>
<div dir="ltr" style="text-align: left;">
XLPE cables are directly laid on solid bed and are used for the voltage upto and including 33 kV.</div>
<div dir="ltr" style="color: blue; text-align: left;">
<u>1.4 Vulcanized India Rubber (VIR)</u></div>
<div dir="ltr" style="text-align: justify;">
This is the most olden insulating material developed during 1880-1930.
The pure rubber is very soft and it can not withstand high
temperatures hence it is 20 to 40% of India rubber mixed with mineral
matter such as zinc oxide, red lead etc. with a little bit of sulphur
in it.</div>
<div dir="ltr" style="text-align: left;">
It has following characteristics,</div>
<div dir="ltr" style="text-align: left;">
1. Good dielectric strength.</div>
<div dir="ltr" style="text-align: left;">
2. Good mechanical strength.</div>
<div dir="ltr" style="text-align: left;">
3. Durable and wear resistant.</div>
<div dir="ltr" style="text-align: left;">
4. Good insulation resistance.</div>
<div dir="ltr" style="text-align: left;">
5. Remain more elastic than pure rubber.</div>
<div dir="ltr" style="text-align: left;">
But it has number of drawbacks such as,</div>
<div dir="ltr" style="text-align: left;">
1. It absorbs moisture, slightly.</div>
<div dir="ltr" style="text-align: left;">
2. It has low melting point.</div>
<div dir="ltr" style="text-align: justify;">
3.
The sulphur content attack the copper conductor and changes the VIR
insulation colour. Hence copper conductors to be used with VIR
insulation must be tinned.</div>
<div dir="ltr" style="text-align: left;">
4. Short span of life.</div>
<div dir="ltr" style="text-align: justify;">
The use of VIR is very limited nowadays and is used for low moderate voltage cable i.e. distribution systems only.</div>
</div>
<ins style="border: none; display: inline-table; height: 280px; margin: 0; padding: 0; position: relative; visibility: visible; width: 336px;"><ins id="aswift_1_anchor" style="border: none; display: block; height: 280px; margin: 0; padding: 0; position: relative; visibility: visible; width: 336px;"></ins></ins>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3691402397720851976.post-16996414684827424922013-04-20T23:09:00.000-07:002013-04-20T23:09:05.149-07:00Gas Pressure Cables In case of as pressure
cables, an inert as like nitrogen at high pressure is introduced. lead
sheath and dielectric. The pressure is about 12 to 15 atmospheres. Due
to such a high pressure there is a radial compression due to which the
ionization is totally eliminated. The working power factors of such
cables is also high.<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhX0FkhVVIdCOA2o9ctdWtfbmNfrfx48N2nh6a4T6TJjOJRd1rbiYzYZabzk0dRnbQbLQjtLmT8NdAUVaC1b1AekXA7Iw5LkVuasXwhDis8rxX_LM04DLvgdUoRjLS01I9eGxmrKjMFypk/s1600/ABB18.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhX0FkhVVIdCOA2o9ctdWtfbmNfrfx48N2nh6a4T6TJjOJRd1rbiYzYZabzk0dRnbQbLQjtLmT8NdAUVaC1b1AekXA7Iw5LkVuasXwhDis8rxX_LM04DLvgdUoRjLS01I9eGxmrKjMFypk/s1600/ABB18.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 1 Gas pressure cables</b></td></tr>
</tbody></table>
<div dir="ltr" style="text-align: justify;">
<br />
</div>
<div dir="ltr" style="text-align: justify;">
The Fig. 1
shows the section of a gas pressure cable. The cable id triangular in
shape and installed in the steel pipe. The pipe is filled with the
nitrogen at 12 to 15 atmospheric pressure. The remaining construction is
similar to that of solid type cable but the thickness of lead sheath is
75% of that of solid type cable. There is no bedding and serving. The
pressure cable was firstly designed by Hochstadter, Vogel and Bownden.</div>
<div dir="ltr" style="text-align: justify;">
The triangle shape lead sheath acts as a pressure membrane. The shape
reduces the weight and provides the low thermal resistance. The high
pressure creates the radial compression to close any voids. The steel
pipe is coated with a point to avoid corrosion. </div>
<div dir="ltr" style="text-align: justify;">
During heating, the cable compound expands and a sheath with acts as a
membrane becomes circular in such a case. When cable cools down the gas
pressure acting via sheath forces compound to come back to the
noncircular normal shape. Due to good thermal characteristics, fire
quenching property and high dielectric strength, the gas SF6 is also
used in such cables.</div>
<div dir="ltr" style="color: blue; text-align: left;">
<u> </u></div>
<div dir="ltr" style="color: blue; text-align: left;">
<b><u>1.1 Advantages</u></b></div>
<div dir="ltr" style="text-align: left;">
</div>
<div dir="ltr" style="text-align: left;">
The various advantages of gas pressure cables are,</div>
<div dir="ltr" style="text-align: justify;">
</div>
<div dir="ltr" style="text-align: justify;">
1.
Gas pressure cables can carry 1.5 times the normal load current and
can withstand double the voltage. Hence such cables can be used for
ultra high voltage (UHV) levels. </div>
<div dir="ltr" style="text-align: left;">
2. Maintenance cost is small.</div>
<div dir="ltr" style="text-align: left;">
3. The nitrogen in the steel tube, helps in quenching any fire or flame.</div>
<div dir="ltr" style="text-align: left;">
4. No reservoir or tanks required.</div>
<div dir="ltr" style="text-align: left;">
5. The power factor is improved.</div>
<div dir="ltr" style="text-align: left;">
6. The steel tubes used make the cable laying easy.</div>
<div dir="ltr" style="text-align: left;">
7. The ionization and possibility of voids is completely eliminated.</div>
<div dir="ltr" style="text-align: left;">
</div>
<div dir="ltr" style="text-align: left;">
The only disadvantages of this type of cables is very high initial cost.</div>
Unknownnoreply@blogger.com1tag:blogger.com,1999:blog-3691402397720851976.post-21339297785693535542013-04-20T23:07:00.000-07:002013-04-20T23:07:07.278-07:00Oil filled Cables In case of oil filled
cables, the channels or ducts are provided within or adjacent to the
cores, through which oil under pressure is circulated.<br />
<div dir="ltr" style="text-align: justify;">
The Fig. 1 shows the construction of single core oil filed cable. It
consists of concentric stranded conductors but built around a hollow
cylindrical steel spiral core. This hollow core acts as a channel for
the oil. The oil channel is filled in a factory and the pressure is
maintained in the oil by connecting the oil channel to the tanks which
are placed at the suitable distances along the path of the cable.</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhXeZMOKb6qFSeSJTo5t3fWXKCxS_1xpmU7ilu-xfFgThxjYnh3NiB1fadJB_i9WZ2ya2mDtYaE5Aml6A_tdfYcbBmoGwD2Qkf-GwoYlQzcJgbfNOhc-qdP59B3vZNzoKReVUhNmxMIke8/s1600/ABB15.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhXeZMOKb6qFSeSJTo5t3fWXKCxS_1xpmU7ilu-xfFgThxjYnh3NiB1fadJB_i9WZ2ya2mDtYaE5Aml6A_tdfYcbBmoGwD2Qkf-GwoYlQzcJgbfNOhc-qdP59B3vZNzoKReVUhNmxMIke8/s1600/ABB15.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 1 Conductor channel single core oil filled cable</b></td></tr>
</tbody></table>
<div dir="ltr" style="text-align: justify;">
<br />
</div>
<div dir="ltr" style="text-align: justify;">
The oil
pressure compresses the paper insulation, eliminating the possibility of
formation of voids. When cable is heated, the oil expands but expanded
oil is collected in the tank. While when cable is cooled, extra oil is
supplied by the tank to maintain the oil pressure. In this type of cable
the oil channel is within the conductor, hence it is called single core
conductor channel oil filled cables. The other construction of the
cable is similar to that of solid type cables.</div>
<div dir="ltr" style="text-align: justify;">
Another type of single core oil filled cable is the sheath channel oil
filled cable. In this type, the conductor is solid with a paper
insulation. While the oil ducts are provided between the dielectric and
the lead sheath.</div>
<div dir="ltr" style="text-align: justify;">
The construction of sheath channel oil filled cable is shown in the
Fig. 2. The laying of such cables must be done very carefully.</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEij2VUbE7OJo9hb4jG6MVEu3bHrhf4o7cHg95Z9Io1l_HprPLW1M4A_6iEyWTGTcVGD2F4RgRr7p9fe49dLVgxTV0rCPULjVnKCuoiGlDRrNb6jg89dTwF1xv06oq3upg3mDJ8BYBnEPKM/s1600/ABB16.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEij2VUbE7OJo9hb4jG6MVEu3bHrhf4o7cHg95Z9Io1l_HprPLW1M4A_6iEyWTGTcVGD2F4RgRr7p9fe49dLVgxTV0rCPULjVnKCuoiGlDRrNb6jg89dTwF1xv06oq3upg3mDJ8BYBnEPKM/s1600/ABB16.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 2 Sheath channel single core oil filled cable </b></td></tr>
</tbody></table>
<div dir="ltr" style="text-align: justify;">
<br />
</div>
<div dir="ltr" style="text-align: justify;">
The three core
oil filled cables use the shielded type construction. The oil channels
are located in the spaces which are normally occupied by the filler
material. The three oil channels are of preforated metal ribbon tubing.
All the channels are at earth potential. The construction is shown in
the Fig. 3. As the pressure tanks are required all all along the route
of these cables, the lengths of these cables are limited. Leakage of oil
is another serious problem associated with these cables. Automatic
signalling units are located to indicate the fall in oil pressure in any
of the phases.</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg7JhSYK2lmVO1qaUDCHWpXZw8j9iRrqa8mILTE-eP_QsmvJPbl7g0tJ3ktj9ruGS_udJ6mC_oGtE-9xTf0UqDXWRSKda5LL3DceZ3ZkPRfjSee-Iz1D-nTQ_bFz3EP8sT2Y-oXJBJCB34/s1600/ABB17.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg7JhSYK2lmVO1qaUDCHWpXZw8j9iRrqa8mILTE-eP_QsmvJPbl7g0tJ3ktj9ruGS_udJ6mC_oGtE-9xTf0UqDXWRSKda5LL3DceZ3ZkPRfjSee-Iz1D-nTQ_bFz3EP8sT2Y-oXJBJCB34/s1600/ABB17.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 3 </b> <b>Oil filled three core cable</b></td></tr>
</tbody></table>
<div dir="ltr" style="text-align: justify;">
<br />
</div>
<div dir="ltr" style="color: blue; text-align: left;">
<u>1.1 Advantages </u></div>
<div dir="ltr" style="text-align: left;">
The various advantages of oil filled cables are,</div>
<div dir="ltr" style="text-align: left;">
1. The thickness of insulation required is less hence smaller in size and weight.</div>
<div dir="ltr" style="text-align: left;">
2. The thermal resistance is less hence current carrying capacity is more.</div>
<div dir="ltr" style="text-align: left;">
3. The possibility of voids is completely eliminated.</div>
<div dir="ltr" style="text-align: left;">
4. The allowable temperature range is more than solid type cables.</div>
<div dir="ltr" style="text-align: justify;">
5.
Reduced possibility of earth fault. This is because in case of any
defect in lead sheath, oil leakage starts, which can be noticed before
earth fault occurs.</div>
<div dir="ltr" style="text-align: left;">
6. Perfect impregnation is possible.</div>
<div dir="ltr" style="color: blue; text-align: left;">
<u>1.2 Disadvantages</u></div>
<div dir="ltr" style="text-align: left;">
The disadvantages of oil filled cables are,</div>
<div dir="ltr" style="text-align: left;">
1. The initial cost is very high.</div>
<div dir="ltr" style="text-align: left;">
2. The long length are not possible.</div>
<div dir="ltr" style="text-align: left;">
3. The oil leakage is serious problem hence automatic signalling equipment is necessary.</div>
<div dir="ltr" style="text-align: left;">
4. The laying of cable is difficult and must be done very carefully.</div>
5. Maintenance of the cables is also complicated.<br />
Unknownnoreply@blogger.com2tag:blogger.com,1999:blog-3691402397720851976.post-81429139945835519992013-04-20T23:05:00.002-07:002013-04-20T23:12:10.243-07:00Super Tension (S.T.) Cables In solid type cables
separate arrangement for avoiding void formation and increasing
dielectric strength is not provided. Hence those cables are used maximum
upto 66 kV level. The S.T. cables are intended for 132 kV to 275 kV
voltage levels.<br />
<div dir="ltr" style="text-align: left;">
<br />
In such cables, the following methods are specially used to eliminate the possibility of void formation :</div>
<div dir="ltr" style="text-align: justify;">
<br />
1.
Instead of solid type insulation, low viscosity oils under pressure is
used for impregnation. The channels are used for oil circulation and oil
is always kept under pressure. The pressure eliminates completely, the
formation of voids.</div>
<div dir="ltr" style="text-align: left;">
2. Using inert gas at high pressure in between the lead sheath and dielectric.</div>
<div dir="ltr" style="text-align: left;">
<br />
Such cables using oil or gas under pressure are called pressure cables and are of two types,</div>
<div dir="ltr" style="text-align: left;">
<br />
<span style="color: blue;">A) Oil filled cables B) Gas Pressure cables</span><br />
<br />
Related posts :<br />
- <b><a href="http://electriciantheory.blogspot.in/2013/04/oil-filled-cables.html" target="_blank">Oil filled Cables </a></b><br />
- <b><a href="http://electriciantheory.blogspot.in/2013/04/gas-pressure-cables.html" target="_blank">Gas Pressure Cables</a></b></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3691402397720851976.post-83036162741701573672013-04-20T23:00:00.001-07:002013-04-20T23:04:41.729-07:00Screened Type Cables These
are used for the voltage levels of 22 kV and 33 kV. The two types of
screened cables are 1. H type cables and 2. S.L. type cables.<br />
<div dir="ltr" style="text-align: left;">
<span style="background-color: white;"><span style="color: blue;"><b><u>1.1 H-Type Cables</u></b></span></span></div>
<div dir="ltr" style="text-align: left;">
<div style="text-align: justify;">
The cable is designed by M.Hochstadter and hence the name given to it
is H-Type cable. There is no paper belt in this type of cable. Each
conductor in this cable is insulated with a paper, covered with a
metallic screen which is generally an aluminium foil. The construction
is shown in the Fig. 1. The metallic screen touches each other. Instead
of paper belt, the three cores are wrapped with a conducting belt which
is usually copper woven fabric tape. Then there is lead sheath. The
conducting belt is in electrical contact with the metallic screen and
lead sheath. After lead sheath there are layers of bedding, armoring and
serving. The metallic screen helps to completely impregnate the cable
which avoids the possibility of formation of voids and spaces. The
conducting belt, the three metallic screens and lead sheath are at earth
potential, due to which electrical stresses are radial in nature. This
keeps the dielectric losses to minimum. Another advantage of metallic
screens is increase in the heat dissipation which reduces the sheath
losses. Due to these advantages, current carrying capacity of these
cables increase. In special cases, the use of these cables cable
extended upto the 66 kV level.</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj7igJiylTeAdaNpDOtUuTxsAGI6SjVB_AFYbgSh2RBTEMdm6K1QBjB4HWaBgxj71tRkYyhcUZNIkSRaoVya9V1vbS-8qnIRlbw85Lxzz2HuygMOW4kzxY6mTY3fEy-n9s4FYzThjlDpoE/s1600/ABB13.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj7igJiylTeAdaNpDOtUuTxsAGI6SjVB_AFYbgSh2RBTEMdm6K1QBjB4HWaBgxj71tRkYyhcUZNIkSRaoVya9V1vbS-8qnIRlbw85Lxzz2HuygMOW4kzxY6mTY3fEy-n9s4FYzThjlDpoE/s1600/ABB13.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>H-type cable</b></td></tr>
</tbody></table>
<div style="text-align: justify;">
<br /></div>
<div>
<span style="background-color: white;"><span style="color: blue;"><b><u>1.2 S.L. Cables</u></b></span></span></div>
<div style="text-align: justify;">
The name S.L. stands for separate lead screened cables. In this cable,
each core is insulated with impregnated paper and each one is then
covered by separate lead sheath. Then there is a cotton tape covering
the three cores together using a paper filler material. Then there are
the layers of armouring and serving. the difference between H-type and
S.L. type cable is that in S.L. type a common lead sheath covering all
the three cores is absent while each core is provided with separate lead
sheath. This allows bending of the cables as per the requirement.</div>
<div style="text-align: justify;">
</div>
<div style="text-align: justify;">
The
construction of S.L.type cable is shown in the Fig. 2.</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjWManq9EazRLaGa-yNukcFedCMHUmI1zvI7Ng7SGNXV5IjViCn3mPkgjZPh_hyphenhyphenGq4GCwpqEjZ8SNtM9EtOeLOV_5b20VzSJx61G_zB1vFlwSc3XrhPP6S2WLt8SDyZZDighQk6uDf_HT4/s1600/ABB14.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjWManq9EazRLaGa-yNukcFedCMHUmI1zvI7Ng7SGNXV5IjViCn3mPkgjZPh_hyphenhyphenGq4GCwpqEjZ8SNtM9EtOeLOV_5b20VzSJx61G_zB1vFlwSc3XrhPP6S2WLt8SDyZZDighQk6uDf_HT4/s1600/ABB14.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>S.L. type cable</b></td></tr>
</tbody></table>
<div style="text-align: justify;">
The three cores in this type of cable are as good as three separate single core cables.</div>
</div>
<div dir="ltr" style="text-align: left;">
<span style="color: blue;"><span style="background-color: white;"><u><b>The various advantages of S.L. type cable are,</b></u></span></span><br />
</div>
<div dir="ltr" style="text-align: left;">
1. Due to lead individual lead sheath, core to core fault possibility gets minimised.<br />
2. The electrical stresses are radial in nature.<br />
3. Due to absence of overall lead sheath, bending of cable is easy.<br />
<div style="text-align: justify;">
4. The dielectric which gets
subjected to electric stresses is paper which is homogeneous hence there
is no possibility of formation of voids.</div>
5. Metal sheath increases the heat dissipation which increases the current carrying capacity.<br />
</div>
<div dir="ltr" style="text-align: left;">
A combination of H-type and S.L. type cable called H.S.L cable also can be used.<br />
<b><i><span style="background-color: lime;"><br /></span></i></b></div>
<div dir="ltr" style="text-align: left;">
<span style="background-color: white;"><span style="color: blue;"><b><u>The lamination of screened cables which are also called solid type cables are</u><i>,</i></b></span></span><br />
</div>
<div dir="ltr" style="text-align: left;">
1. It uses solid insulation only like paper. When the conductor
temperature increases, the paper gets expanded. This eventually
stretches the lead sheath.<br />
2. When the load on the cable decreases, it cools down and there is
contraction of lead sheath. Due to this air may be drawn into the cable
forming voids. This deteriorates the cable insulation.<br />
3. Moisture may be drawn in alongwith the air which deteriorates the dielectric strength of dielectric.<br />
4. Mechanical shock can cause voids. The breakdown strength of voids in
much less than insulation. Hence voids can cause permanent damage to the
cables.<br />
<br /></div>
<ins style="border: none; display: inline-table; height: 280px; margin: 0; padding: 0; position: relative; visibility: visible; width: 336px;"><ins id="aswift_1_anchor" style="border: none; display: block; height: 280px; margin: 0; padding: 0; position: relative; visibility: visible; width: 336px;"></ins></ins>Unknownnoreply@blogger.com1tag:blogger.com,1999:blog-3691402397720851976.post-74842902486814519302013-04-20T22:49:00.003-07:002013-04-20T22:49:47.495-07:00Belted cables As mentioned earlier, these are used for the voltage levels upto 11 KV. The construction of belted cable is shown in the Fig.1.<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhudykf9XXm8Lw-NTMXvc4UUZNcN53AYVOstBGNN0HYe2Td2YPJgciiVKNQqXldtTlS2UXAD3c0biV85ROVrmwOgWniMBjynYxNTDJrriJ0i1ZfmdRZHy_vpXd0_hqv_8j1D24VNYbDl60/s1600/ABB15.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" height="168" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhudykf9XXm8Lw-NTMXvc4UUZNcN53AYVOstBGNN0HYe2Td2YPJgciiVKNQqXldtTlS2UXAD3c0biV85ROVrmwOgWniMBjynYxNTDJrriJ0i1ZfmdRZHy_vpXd0_hqv_8j1D24VNYbDl60/s320/ABB15.jpeg" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 1 Belted 3 core cable</b></td></tr>
</tbody></table>
<div dir="ltr" style="text-align: justify;">
<br />
</div>
<div dir="ltr" style="text-align: justify;">
The cores are
not circular in shape. The core are insulated from each other by use of
impregnated paper. The three cores are grouped together and belted with
the help of a paper belt. The gaps are filled with fibrous material like
jute. This gives circular cross-sectional shape to the cable. The belt
is covered with lead sheath which protects cable from moisture and also
gives mechanical strength. The lead sheath is finally covered by jute
like fibrous compounded material.</div>
<div dir="ltr" style="text-align: justify;">
The electrical field in single core cable is radial while it is
tangential in case of three core cables. Hence the insulation is
subjected to tangential electrical stresses rather than radial one. The
paper has good radial strength but not tangential strength. Similarly
paper resistance along the radius is much larger than resistance along
tangential path. The same is true for dielectric strength also. The
fibrous material is also subjected to the tangential electrical
stresses, for which, the material is weak. Hence under high voltage
cases, the cumulative effect of tangential electrical stresses is to
from space inside the cable due to leakage currents. Such air spaces
formed inside the insulation is called void formation. This void
formation is dangerous because under high voltage, spaces are ionized
which deteriorates the insulation which may lead to the breakdown of the
insulation. Hence the belted cables are not used for the high voltage
levels. Another disadvantage of the belted cable is large diameter of
paper belt. Due to this, wrinkles are formed and gaps may be developed
if the cable is bended. To overcome all theses difficulties, the
screened type cables are used.</div>
<ins style="border: none; display: inline-table; height: 280px; margin: 0; padding: 0; position: relative; visibility: visible; width: 336px;"><ins id="aswift_1_anchor" style="border: none; display: block; height: 280px; margin: 0; padding: 0; position: relative; visibility: visible; width: 336px;"></ins></ins>Unknownnoreply@blogger.com2tag:blogger.com,1999:blog-3691402397720851976.post-8849630035252130512013-04-20T22:48:00.001-07:002013-04-20T22:48:36.621-07:00Types of Cables The type of a cable is
basically decided based on the voltage level for which it is
manufactured and the material used for the insulation such as paper,
cotton, rubber etc. The classification of cables according to the
voltage level is,<br />
<div dir="ltr" style="text-align: justify;">
1- <b>Low tension cables</b>
(L.T. cables) : These are used for the voltage levels upto 6.6 KV. The
electrostatic stresses in L.T. cables are not severe hence no special
construction is used for L.T. cables. The paper is used as an insulation
in theses cables. Sometimes resin in also used which increases the
viscosity and helps to prevent drainage.<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhqz-QkW-EM4Ys-JAaPiVHTp4QBHOIiddtKpmdvIOGULOvLo8Zq0FmQV12-Nvrf8kZ68xE9SXQjXDZYv-rIZV_eqYpZrrIQAePi7JeKzNJBVYvY13aWW3alWEnrLXdtvGMN2L_NBTD5Nrw/s1600/ABB14.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhqz-QkW-EM4Ys-JAaPiVHTp4QBHOIiddtKpmdvIOGULOvLo8Zq0FmQV12-Nvrf8kZ68xE9SXQjXDZYv-rIZV_eqYpZrrIQAePi7JeKzNJBVYvY13aWW3alWEnrLXdtvGMN2L_NBTD5Nrw/s1600/ABB14.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 1 Single core L.T. cable</b></td></tr>
</tbody></table>
</div>
<div style="text-align: justify;">
The Fig.1 shows the cross-section of a single core L.T. cable. It
consists of a circular core of stranded copper or aluminium. The
conductor is insulated by impregnated paper. Over the paper insulation,
the lead sheath is provided. Then a layer of compounded fibrous material
is provided. Then armouring is provided and finally covered again with a
layer of fibrous compounded material.</div>
<div style="text-align: justify;">
Many a times, L.T. cables are not provided with armouring, to avoid
excessive sheath losses. The simple construction and the availability of
more copper section are the advantages of L.T. single core cable.</div>
<div style="text-align: justify;">
2- <b>Medium and high tension cables</b>
(H.T. cables) : The three phase medium and H.T. cables are three core
cables. For voltages upto 66 KV, the three core cables i.e. multicore
cables are used. These cables are classified as,</div>
a. H.T. cables upto 11 KV level which are belted type.<br />
b. Super tension (S.T.) cables for 22 KV and 33 KV levels which are screened cables.<br />
c. Extra high tension (E.H.T.) cables for voltage levels from 33 KV to 66 KV which are pressure cables.Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3691402397720851976.post-51127365127027203892013-04-20T22:47:00.000-07:002013-04-20T22:47:16.997-07:00General Construction of a Cable The Fig. 1 shows the general
construction of a cable. The cable shown is a single conductors
underground cable. Its various parts are,<br />
<div dir="ltr" style="text-align: justify;">
<span style="color: blue;"><b>1- Conductor or core</b></span>
: This section consists of single conductor or more than one
conductors. The conductors are also called cores. A cable with three
conductors is called three core cable. The conductors used are aluminium
or annealed copper. The conductors are stranded conductors in order to
provide flexibility to the cable.</div>
<div dir="ltr" style="text-align: justify;">
<b><span style="color: blue;">2- Insulation
: </span></b>Each conductors or core is covered by insulation of proper thickness.
The commonly used insulating materials are varnished cambric,
vulcanized bitumen and impregnated paper.</div>
<div dir="ltr" style="text-align: justify;">
<span style="color: blue;"><b>3- Metallic sheath :</b></span>
The insulated conductors are covered by lead sheath or aluminium
sheath. This provides the mechanical protection but mainly restricts
moisture and other gases to reach to the insulation.</div>
<div dir="ltr" style="text-align: justify;">
<span style="color: blue;"><b>4- Bedding
:</b></span> The metallic sheath is covered by by another layer called bedding.
The bedding consists of paper tape compounded with a fibrous material
like jute strands or hessian tape. The purpose of bedding is to protect
the metallic sheath from corrosion and from mechanical injury resulting
due to aemouring.</div>
<div dir="ltr" style="text-align: justify;">
<b><span style="color: blue;">5- Armouring :</span></b> This layer consists of the layer of galvanized steel wires which provide protection to the cable from the mechanical injury.</div>
<div dir="ltr" style="text-align: justify;">
<span style="color: blue;"><b>6- Serving
:</b></span> The last layer above the armouring is serving. It is a layer of
fibrous material like jute cloth which protects the armouring from the
atmospheric conditions.</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgz0BO0Ls-iRn47lXySEdOHMJogalXI6FTj09dWG21GoL-gecKp76bKlUUqbk6OIpjiARz3irvyNkS2S41b_qwdAvr7JkhKjZbHm2vCMquG4dS9D-jdGEDzhPrDo_P1KhrEK5Ftgzbq0Lk/s1600/ABB13.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgz0BO0Ls-iRn47lXySEdOHMJogalXI6FTj09dWG21GoL-gecKp76bKlUUqbk6OIpjiARz3irvyNkS2S41b_qwdAvr7JkhKjZbHm2vCMquG4dS9D-jdGEDzhPrDo_P1KhrEK5Ftgzbq0Lk/s1600/ABB13.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 1 General construction of a cable</b></td></tr>
</tbody></table>
<div dir="ltr" style="text-align: left;">
<br />
</div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3691402397720851976.post-16234666294679617292013-04-20T22:44:00.007-07:002013-04-20T22:44:57.940-07:00Requirements of the Cables An underground cable can be
defined as the group of individually insulated one or more conductors
which is put together and finally provided with number of layers of
insulations to give proper mechanical support.<br />
<div dir="ltr" style="text-align: justify;">
The conductors used in the cables are usually aluminium or annealed
copper while the insulation is commonly PVC or other chemical
compositions. Many types of cables are available depending upon the
nature of conductors, number of conductors, types of insulation used
etc. The basic necessary requirements of the cables are,</div>
<div dir="ltr" style="text-align: justify;">
</div>
<div dir="ltr" style="text-align: justify;">
1-
The size of the conductors used must be such that it should be carry
the specified load without overheating and keeping the voltage drop well
within the permissible limits.</div>
<div dir="ltr" style="text-align: justify;">
</div>
<div dir="ltr" style="text-align: justify;">
2-
As the voltage level for which cables are designed, the insulation
thickness must be proper so as to provide high degree of safety and the
reliability.</div>
<div dir="ltr" style="text-align: justify;">
</div>
<div dir="ltr" style="text-align: justify;">
3- The
cables must be surrounded by number of layers of an additional
insulation so as to give proper mechanical strength and protection. Thus
the cables can withstand the rough use at the time of laying them.</div>
<div dir="ltr" style="text-align: justify;">
</div>
<div dir="ltr" style="text-align: justify;">
4-
The material used in the manufacturing of cables must be such that
there is complete chemical and physical stability throughout.</div>
<div dir="ltr" style="text-align: left;">
<br />
</div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3691402397720851976.post-30812269000462531682013-04-20T22:43:00.004-07:002013-04-20T22:43:38.179-07:00Cables <b><span style="color: blue;"><u>1.1 Introduction</u></span></b><br />
<div dir="ltr" style="text-align: justify;">
The transmission and distribution of an electrical power can be with
the help of overhead transmission lines or by underground cables. It has
been mentioned that in thickly populated are like towns and cities, the
use of overhead lines is not practicable. In such cases electrical
energy is transmitted and distributed with the help of underground
cables. In its basic form, an underground cable is a conductor provided
with proper insulation. As the voltage level increases, the cost of the
insulation increases rapidly and thus the use of underground cables is
restricted to low and medium voltage distribution.</div>
<div dir="ltr" style="text-align: justify;">
<br />
</div>
<div dir="ltr" style="color: blue; text-align: left;">
<b><u>1.2 Comparison of Underground cables and Overhead lines</u></b></div>
<div dir="ltr" style="text-align: left;">
Compared to overhead lines, the underground cables have the following advantages,</div>
<div dir="ltr" style="text-align: justify;">
1.
It ensures non-interrupted continuity of supply. The possible supply
interruptions due to lighting, storms and other weather conditions are
eliminated because of underground cables.</div>
<div dir="ltr" style="text-align: left;">
2. It requires less maintenance.</div>
<div dir="ltr" style="text-align: justify;">
3. The accidents caused due to breakage of overhead line conductors are eliminated due to use of underground cables.</div>
<div dir="ltr" style="text-align: left;">
4. The voltage drop in the underground cables is less.</div>
<div dir="ltr" style="text-align: left;">
5. The life of underground cables is long compared to overhead lines.</div>
<div dir="ltr" style="text-align: left;">
6- The beauty of cities and town get maintained due to underground network of cables.</div>
<div dir="ltr" style="text-align: justify;">
7-
The overhead lines use bars conductors which is unsafe in thickly
populated area. Hence from safety point of view, the underground cables
are more advantageous.</div>
<div dir="ltr" style="text-align: justify;">
The only drawbacks of underground cables are the extremely high initial
cost and insulation problems at high voltages. In India, the big cities
have adopted the system of underground cables for the transmission and
distribution.</div>
<div dir="ltr" style="text-align: justify;">
Thus the use of underground cables is mainly for the distribution of an
electrical power at low and medium voltages. Its use is almost
compulsory at the location where use of overhead lines is not
practicable due to the safety reasons such as congested urban area,
crossing of wide roads, near gas plants and refineries, near substation
etc.</div>
<div dir="ltr" style="text-align: left;">
Still the overhead lines also have some advantages compared to the underground cables which are,</div>
<div dir="ltr" style="text-align: left;">
1- Long distance transmission is possibly by the overhead lines.</div>
<div dir="ltr" style="text-align: left;">
2- The conductors in overhead lines is less expensive.</div>
<div dir="ltr" style="text-align: justify;">
3- The size of the conductor in overhead lines is less than underground cables due to good heat dissipation in overhead lines.</div>
<div dir="ltr" style="text-align: justify;">
4-
The insulation cost is very less as the air itself acts as an
insulation between the conductors. The gas or oil is not required for
overhead lines. For high voltage levels, spacing in air can be easily
adjusted in case of overhead lines to obtain proper insulation.</div>
<div dir="ltr" style="text-align: justify;">
5- The erection cost is much less for the overhead lines. The underground cable laying is difficult and complicated.</div>
<div dir="ltr" style="text-align: justify;">
Depending upon the situation and the requirements, the underground cables or overhead transmission system can be used. </div>
<ins style="border: none; display: inline-table; height: 280px; margin: 0; padding: 0; position: relative; visibility: visible; width: 336px;"><ins id="aswift_1_anchor" style="border: none; display: block; height: 280px; margin: 0; padding: 0; position: relative; visibility: visible; width: 336px;"></ins></ins>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3691402397720851976.post-88841011525081924392013-04-20T22:39:00.001-07:002013-04-20T22:39:30.202-07:00Supply Systems Introduction The energy is neither be
created nor be destroyed but it can be converted from one form to
another. The generation of an electrical energy is nothing but the
conversion of various other forms of energy into an electrical energy.
The various energy sources which are used to generate an electrical
energy on the large scale are steam obtained by burning coal, oil,
natural gas, water stored in dams, diesel oil, nuclear power and other
nonconventional energy sources. The electrical power is generated in
bulk at the generating stations which are also called power stations.
Depending up on the source of energy used, these stations are called
thermal power station, hydroelectric power station, diesel power
station, nuclear power station etc.
<br />
<div dir="ltr" style="text-align: left;">
</div>
<div dir="ltr" style="text-align: justify;">
This generated electrical energy is demanded by the consumers.
Hence the generated electrical power is to be supplied to the consumers.
Generally the power stations are located too far away from the town and
cities where electrical energy is demanded. Hence there exists a large
network of conductors between the power stations and the consumers. </div>
<div dir="ltr" style="text-align: justify;">
<b>This
network is broadly classified into two parts.</b></div>
<div dir="ltr" style="text-align: left;">
</div>
<div dir="ltr" style="text-align: left;">
<span style="color: blue;">1. Transmission 2. Distribution</span></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3691402397720851976.post-63161031261209254502013-04-20T22:36:00.002-07:002013-04-20T22:36:31.703-07:00 A typical Transmission and Distribution Scheme The
flow of electrical power from the generating station to the consumer is
called an electrical power system or electrical supply system. It
consists of the following important components :
<br />
<div dir="ltr" style="text-align: left;">
<div style="text-align: justify;">
</div>
<div style="text-align: justify;">
1. Generating station 2. Transmission network 3. Distribution Network</div>
</div>
<div dir="ltr" style="text-align: left;">
<div style="text-align: justify;">
</div>
<div style="text-align: justify;">
All these important networks are connected with the help of
conductors and various step up and step down transformers. A typical
transmission and distribution scheme is shown in the Fig. 1.</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhGcVr2EBAlbZzauf3sjLHCpLWF4BZek7NsQiwQjv7_R6ZHI4GKZBMyGd6J0EepinvHGOp-EuyGDdo_3dZNWhNEDP7C-hw2IY20HvZDCdQNoCjY4eeTzFi41l-UEuwBsdv_UlgYbihoMiBF/s1600/ccc116.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhGcVr2EBAlbZzauf3sjLHCpLWF4BZek7NsQiwQjv7_R6ZHI4GKZBMyGd6J0EepinvHGOp-EuyGDdo_3dZNWhNEDP7C-hw2IY20HvZDCdQNoCjY4eeTzFi41l-UEuwBsdv_UlgYbihoMiBF/s1600/ccc116.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 1 Schematic Representation of a typical transmission distribution scheme</b></td></tr>
</tbody></table>
<div style="text-align: justify;">
</div>
</div>
<div dir="ltr" style="text-align: left;">
<div style="text-align: justify;">
A scheme shows a generating station which is located too far away
from cities and towns. It is generating an electrical power at 11 Kv.
It is required to increase this level for the transmission purpose.
Hence a step up transformer is used which steps up the voltage level to
220 Kv. This level may be 132 Kv, 220 Kv or more as per the requirement.</div>
</div>
<div dir="ltr" style="text-align: left;">
<div style="text-align: justify;">
Then with the help of transmission lines and the towers, the
power is transmitted at very long distances. Design of the transmission
lines is based on the factors like transmission voltage level, constants
like resistance, reactance of the lines, line performance, interference
with the neighbouring circuits etc. Its mechanical features are
strength of the supports, sag calculations, tension etc. Transmission of
power by the overhead lines is very much cheaper. Similarly the repairs
also can be carried out comparatively more easily. The transmission is
generally along with additional lines in parallel. These lines are
called duplicate lines. Thus two sets of three phase lines work in
parallel. This ensures the continuity during maintenance and also can be
used to satisfy future demand. The power is then transmitted to the
receiving station via step down transformer. This transformer is 220/33
kv or 220/22 Kv transformer.</div>
</div>
<div dir="ltr" style="text-align: left;">
<div style="text-align: justify;">
The power is then transmitted to the substations. A substation
consists of a step down transformer of rating 33 KV to 6.6 Kv or 3.3 KV.
The transfer of power from receiving station to the substation is with
the help of conductors called feeders. This is called secondary
transmission.</div>
</div>
<div dir="ltr" style="text-align: left;">
<div style="text-align: justify;">
From the substations, power is distributed to the local
distribution centers with the help of distributors. Sometimes for bulk
loads like factories and industries, the distributors transfer power
directly. For the light loads, there are distribution centers consisting
of distribution transformers which step down the voltage level to 230 V
or 400 V. This is called primary distribution. In the crowded areas
like cities, overhead system of bare conductors is not practicable. In
such cases insulated conductors are used on the form of underground
cables, to give supply to the consumers. These cables are called service
mains. This is called secondary distribution.</div>
<div style="text-align: justify;">
This is the complete flow of an electrical power from the generating station to the consumer permises.</div>
<div style="text-align: justify;">
Let us study the line diagram of such a typical scheme of
transmission and distribution and discuss the various components and
voltage levels at the various stages in detail. The Fig. 2 shows the
line diagram of a typical transmission and distribution scheme.</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhXjQDEfXma20YiKVNfPdpzDa8gH77XtAlT16pzy7EbePE8YosIRsvgFHWCNLeIXUtSav-BfWDwbeRV0q6shlkTYXrqdRBVfizfB6lnnlsQVpJwNojd_fMgxL3mudZ0hXD8EplSYJ53DVh-/s1600/ccc117.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhXjQDEfXma20YiKVNfPdpzDa8gH77XtAlT16pzy7EbePE8YosIRsvgFHWCNLeIXUtSav-BfWDwbeRV0q6shlkTYXrqdRBVfizfB6lnnlsQVpJwNojd_fMgxL3mudZ0hXD8EplSYJ53DVh-/s1600/ccc117.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 2 Line diagram of a typical transmission distribution scheme</b></td></tr>
</tbody></table>
<div style="text-align: justify;">
</div>
<div style="text-align: justify;">
At the generating station, an electrical power is generated with
the help of three phase alternators running in parallel. In the scheme
shown, the voltage level is 11 KV but the voltage level may be 6.6 KV,
22 KV or 33 KV depending upon the capacity of the generating station.
After the generating station, actual transmission and distribution
starts. The overall scheme can be divided into four sections which are,</div>
<div style="text-align: justify;">
<b> </b></div>
<div style="text-align: justify;">
<b>1. Primary transmission </b>: It is basically with the help of
overhead transmission lines. For the economic aspects, the voltage level
is increased to 132 KV, 220 KV or more, with the help of step up
transformer. Hence this transmission is also called high voltage
transmission. The primary transmission uses 3 phase 3 wire system.</div>
<div style="text-align: justify;">
<b> </b></div>
<div style="text-align: justify;">
<b>2. Secondary Transmission </b>: The primary transmission line
continues via transmission towers till the receiving stations. At the
receiving stations, the voltage level is reduced to 22 KV or 33 KV using
the step down transformer. There can be more than one receiving
stations. Then at reduced voltage level of 22 KV or 33 KV, the power is
then transmitted to various substations using overhead 3 phase 3 wire
system. This is secondary transmission. The conductors used for the
secondary transmission are called feeders.</div>
<div style="text-align: justify;">
<b> </b></div>
<div style="text-align: justify;">
<b>3. Primary Distribution </b>: At the substation the voltage level is
reduced to 6.6 KV, 3.3 KV or 11 KV with the help of step down
transformers. It uses three phase three wire underground system. And the
power is further transmitted to the local distribution centers. This is
primary distribution, also called high voltage distribution. For the
large consumers like factories and industries, the power is directly
transmitted to such loads from a substation. Such bug loads have their
own substations.</div>
<div style="text-align: justify;">
<b> </b></div>
<div style="text-align: justify;">
<b>4. Secondary Distribution : </b>At the local distribution centers,
there are step down distribution transformers. The voltage level of 6.6
KV, 11 KV is further reduced to 400 V using distribution transformers.
Sometimes it may be reduced to 230 V. The power is then transmitted
using distribution, also called low voltage distribution. This uses 3
phase 4 wire system. The voltage between any two lines is 400 V. while
the voltage between any of the three lines and a neutral is 230 V. The
single phase lighting loads are supplied using a line and neutral while
loads like motors are supplied using three phase lines.</div>
<u><span style="color: blue;"> </span></u></div>
<div dir="ltr" style="text-align: left;">
<b><u><span style="color: blue;">1.1 Components of Distribution</span></u></b><br />
</div>
<div dir="ltr" style="text-align: left;">
The distribution scheme consists of following important components :<br />
<div style="text-align: justify;">
<b> </b></div>
<div style="text-align: justify;">
<b>1. Substation : </b>Transmission lines bring the power upto the
substations at a voltage level of 22 KV or 33 KV. At the substation the
level is reduced to 3.3 KV or 6.6 KV. Then using feeders, The power is
given to local distribution centers.</div>
<div style="text-align: justify;">
<b> </b></div>
<div style="text-align: justify;">
<b>2. Local distribution station :</b> It consists of distribution
transformer which steps down the voltage level from 3.3 KV, 6.6 KV to
400 V or 230 V. Then it is distributed further using distributors. This
is also called distribution substation.</div>
<div style="text-align: justify;">
<b> </b></div>
<div style="text-align: justify;">
<b>3. Feeders :</b> These are the conductors which are of large current
carrying capacitor. The feeders connect the substation to the are where
power is to be finally distributed to the consumers. No tapping are
taken from the feeders. The feeder current always remains constant.</div>
<div style="text-align: justify;">
<b> </b></div>
<div style="text-align: justify;">
<b>4. Distributors : </b>These are the conductors used to transfer power
from distribution center to the consumers. From the distributors, the
tappings are taken for the supply to the consumers.</div>
<div style="text-align: justify;">
<b> </b></div>
<div style="text-align: justify;">
<b>5. Service Mains : </b>These are the small cables between the distributors and the actual consumers permises.</div>
<div style="text-align: justify;">
The interconnection of feeders, distributors and service mains is shown in the Fig. 3.</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjDaNFCQlG6LdeRe1MYpr7rsK8S2DWHPsSqmhSJFio0bf52zTwgPrHfpOFBJQr5VQ8A6uT8zJjR8HlcZc31XwbN5CHOZoasM-2QWFfnaR1UJyIlA8IqHKCfBhgrv31kTL1Vzvjx_4hx_ng5/s1600/ccc118.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjDaNFCQlG6LdeRe1MYpr7rsK8S2DWHPsSqmhSJFio0bf52zTwgPrHfpOFBJQr5VQ8A6uT8zJjR8HlcZc31XwbN5CHOZoasM-2QWFfnaR1UJyIlA8IqHKCfBhgrv31kTL1Vzvjx_4hx_ng5/s1600/ccc118.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Fig. 3</td></tr>
</tbody></table>
<div style="text-align: justify;">
</div>
<div style="text-align: justify;">
There is no tapping on feeders. PQ, QR, RS and PS are the
distributors which are supplied are supplied by the feeder. The service
mains are used to supply the consumers from the distributors. Tappings
are taken from the distributors.</div>
<div style="text-align: justify;">
</div>
</div>
<ins style="border: none; display: inline-table; height: 280px; margin: 0; padding: 0; position: relative; visibility: visible; width: 336px;"><ins id="aswift_1_anchor" style="border: none; display: block; height: 280px; margin: 0; padding: 0; position: relative; visibility: visible; width: 336px;"></ins></ins>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-3691402397720851976.post-54849897836061864652013-04-20T22:33:00.001-07:002013-04-20T22:33:24.538-07:00Elements of Power System The
power system is comprised of various elements such as generator,
transformer, transmission lines, bus bars, circuit breakers, isolators
etc. Now we will discuss in brief about these elements.
<br />
<div dir="ltr" style="text-align: left;">
<div style="text-align: justify;">
<u><span style="color: blue;">1.1 Generators</span></u> </div>
<div style="text-align: justify;">
The generator or alternator is the important element of power
system. It is of synchronous type and is driven by turbine thus
converting mechanical energy into electrical energy. The two main parts
of generator are stator and rotor. The stationary part is called stator
or armature consisting of conductors embedded in the slots. The
conductors carry current when load is applied on the generator. The
rotating part or rotor is mounted on the shaft and rotates inside the
stator. The winding on rotor is called field winding. The field winding
is excited by d.c. current. This current produces high m.m.f. The
armature conductors react with the m.m.f. produced by the field winding
and e.m.f. gets induced in the armature winding. The armature conductors
carry current when the load is connected to an alternator. This current
produces its own m.m.f. This m.m.f. interacts with the m.m.f. produced
by the field winding to generate an electromagnetic torque between
stator and rotor.</div>
</div>
<div dir="ltr" style="text-align: left;">
<div style="text-align: justify;">
The d.c. current required for field winding is supplied through
exciter which is nothing but a generator mounted on the same shaft on
which alternator is mount. The separate d.c. source may also be used
sometimes to excite the field windings through brushes bearing o slip
rings.</div>
</div>
<div dir="ltr" style="text-align: left;">
<div style="text-align: justify;">
The generators are driven by prime mover which is normally a
steam or hydraulic turbine. The electromagnetic torque developed in the
generator while delivering power opposes the torque provided by the
prime mover.</div>
</div>
<div dir="ltr" style="text-align: left;">
<div style="text-align: justify;">
With properly designed rotor and proper distribution of stator
windings around the armature, it is possible to get pure sinusoidal
voltage from the generator. This voltage is called no load generated
voltage or generated voltage. The representation of generator is shown
in the Fig. 1.</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgpWnYzUcWe2I3xwgwtXbJndVctCGTBiD7pwamYXAAUcNwSdKq6Aj2R4wlthH9tbkRAr6kLAgsjEb3k8SiBN5VFvID1M51CjcE_ZdLkcUoq0LEEissscGt_xAzUb3D3MSishtbWefYZV6c5/s1600/ccc119.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgpWnYzUcWe2I3xwgwtXbJndVctCGTBiD7pwamYXAAUcNwSdKq6Aj2R4wlthH9tbkRAr6kLAgsjEb3k8SiBN5VFvID1M51CjcE_ZdLkcUoq0LEEissscGt_xAzUb3D3MSishtbWefYZV6c5/s1600/ccc119.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 1 Representation of alternator</b></td></tr>
</tbody></table>
<div style="text-align: justify;">
</div>
</div>
<div dir="ltr" style="text-align: left;">
<div style="text-align: justify;">
<span style="color: blue;"><u>1.2 Transformer</u></span></div>
</div>
<div style="text-align: justify;">
For stepping up or down the system voltage, power transformers
are used in the substations. At generating end, the voltage is only
stepped up for transmission of power while at all the subsequent
substations the voltage is gradually stepped down to reach finally to
working voltage level.</div>
<div style="text-align: justify;">
Instead of using a bank of 3 single phase transformers, a single
three phase transformer is used nowadays. The advantages of using this
transformer is the easiness in its installation and only one three phase
load tap changing mechanism can be used.</div>
<div style="text-align: justify;">
Generally naturally cooled, oil immersed, two winding, three
phase transformers upto the rating of 10 MVA are installed upon lengths
of rails fixed on concrete slabs having foundations 1 to 1.5 m deep. For
more than 10 MVA ratings, forced oil, water cooling and air blast
cooling type may be used. The tap changers are used for regulating the
voltage of transformers.<br />
</div>
<div style="text-align: justify;">
<span style="color: blue;"><u>1.3 Transmission Line</u></span></div>
<div style="text-align: justify;">
The transmission line forms the connecting link between the
generating stations and the distribution systems. It carries the power
generated by generating stations and makes it available for distribution
through distribution network.</div>
<div style="text-align: justify;">
Any electrical transmission line has four major parameters which
are important from the point view of its proper operation. These
parameters are namely resistance, inductance, capacitance and
conductors.</div>
<div style="text-align: justify;">
The resistance and inductance is uniformly distributed along the
line. It forms series impedance. The resistance of a line is responsible
for power loss. It is expected that the resistance of a line should be
as low as possible so that the transmission system will be more
efficient. Due to linkage, the conductor is associated with inductance
which is distributed along the length of the line. For analysis, both
resistance and inductance are assumed to be lumped.</div>
<div style="text-align: justify;">
The capacitance also exists between the conductors and is the
charge on the conductors per unit of potential difference between them.
The conductance between conductors or between conductors and the ground
is due to leakage current at the insulators of overhead lines and
through the insulation of cables. The leakage at conductors is
negligible so the conductance between conductors of an overhead line is
taken as zero. The conductance and capacitance between conductors of a
single phase line or from conductor to neutral of a three phase line
form the shunt admittance.</div>
<div style="text-align: justify;">
Depending upon the length of the transmission line it is
classified as short transmission line, medium transmission line and long
transmission line. For short line, its length is small so capacitance
effects are small and are neglected.<br />
</div>
<div style="text-align: justify;">
<span style="color: blue;"><u>1.4 Bus Bars</u></span></div>
<div style="text-align: justify;">
Bus bars are the common electrical component that connect
electrically number of lines which are operating at the same voltage
directly. These bars are of either copper or aluminium generally of
rectangular cross-section. They can be of other shapes such as round
tubes, round solid bars or square tubes.</div>
<div style="text-align: justify;">
The outdorr bus bars, pipes are used. The pipes are also used for
making connections among different components. The pedestal insulators
support the bus bars and the connections. The equipments and bus bars
are spared out and it requires large space. The clearance remain
constant s the bus bars are rigid.</div>
It has following advantages.<br />
1) The maintenance is easy as bus bars and connections are not very high from ground.<br />
2) As pipe diameter is large, the corona loss is less.<br />
3) Reliability is more than strain type.<br />
Following are its limitations.<br />
1) Larger area is required.<br />
2) It requires comparatively high cost.<br />
<div style="text-align: justify;">
In strain type, bus bars are an overhead system of wires between
two supporting structure and supported by strain type insulators. As per
the size of the conductor, the stringing tension can be limited (500 -
900 kg).</div>
<div style="text-align: justify;">
The advantage of this type is its economy and its recommended presently due to general shortage of aluminium pipes.</div>
<div style="text-align: justify;">
The material used in case of rigid type bus bars is aluminium
pipes. The general sizes of pipes commonly used for voltage are as given
below.</div>
<div style="text-align: center;">
33 kv 40 mm</div>
<div style="text-align: center;">
66 kv 65 mm</div>
<div style="text-align: center;">
132 kv 80 mm</div>
<div style="text-align: center;">
220 kv 80 mm</div>
<div style="text-align: center;">
400 kv 100 mm</div>
<div style="text-align: justify;">
Due to rapid oxidization of aluminium, proper care must be taken
while doing connections. In order to avoid strain of supporting
insulators due to thermal expansion or contraction of pipe, joints
should be provided.</div>
<div style="text-align: justify;">
In case of strain type arrangement, material used is ACSR
(Aluminium conductors with steel reinforcement) and all aluminium
conductors. For high ratings of bus bars bundled conductors are used.
The commonly used sizes are as below.</div>
<div style="text-align: center;">
66 kv 37/2.79 mm ACSR</div>
<div style="text-align: center;">
132 kv 37/4.27 mm ACSR</div>
<div style="text-align: center;">
220 kv 61/3.99 mm ACSR</div>
<div style="text-align: center;">
400 kv 61/7.27 mm ACSR in duplex</div>
<br />
<span style="color: blue;"><u>1.5 Circuit Breakers</u></span><br />
<div style="text-align: justify;">
The circuit breakers are used to open or close a circuit under
normal and faulty conditions. It can be designed in such a way that it
can be manually operated or by remote control under normal conditions
and automatically operated during fault. For automatic operation, relay
circuit is used.</div>
<div style="text-align: justify;">
The circuit breakers are essential as isolators cannot be used to
open a circuit under normal conditions as it has no provision to quench
arc that is produced after opening the line. It has perform following
functions.</div>
i) Full load current is to be carried continuously.<br />
ii) Opening and closing the circuit on no load.<br />
iii) Making and breaking the normal operating current.<br />
iv) Making and breaking the fault currents of magnitude upto which it is designed for.<br />
<div style="text-align: justify;">
Upto 66 KV voltages, bulk oil circuit breaker are used. Voltages
greater than 66 KV, low oil circuit breaker are used. For still high
voltages, air blast, vacuum or SF6 circuit breakers are used.<br />
</div>
<div style="text-align: justify;">
<span style="color: blue;"><u>1.6 Isolators </u></span></div>
<div style="text-align: justify;">
In order to disconnect a part of the power system for maintenance
and repair purposes, isolating switches are used. These are operated
after switching off the load by means of a circuit breaker. The
isolators are connected on the both sides of circuit breakers. Thus to
open isolators, circuit breakers are to be opened first.</div>
<div style="text-align: justify;">
An isolator is essentially a knife switch and is designed to open
a circuit under no load that is lines in which are connected should be
carrying any current.</div>
Use of isolators in a substation is shown in the Fig. 2.<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi6RNPQd6X7VQsVbAmcbp4gZOXzs_U51EhrYtwloAcRRILdGtDXD7XMV4g_zjRwkGlAuertAFVFc4P4AssLmxLQ1-5boFrKAJ32Bh85YP9fZz0RGfx9OStiWi8PFQrnjPcAxViLHCi3uXEk/s1600/ccc120.jpeg" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi6RNPQd6X7VQsVbAmcbp4gZOXzs_U51EhrYtwloAcRRILdGtDXD7XMV4g_zjRwkGlAuertAFVFc4P4AssLmxLQ1-5boFrKAJ32Bh85YP9fZz0RGfx9OStiWi8PFQrnjPcAxViLHCi3uXEk/s1600/ccc120.jpeg" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><b>Fig. 2 Line diagram of substation with use of isolating switches</b></td></tr>
</tbody></table>
<br />
<div style="text-align: justify;">
As shown in thew Fig. 2, there are 5 sections. With the help of
isolators, each section can be disconnected for repair and maintenance.
If it is required to do maintenance in section 4, then the circuit
breaker in that section is to be opened first and then open the isolator
3 and 4. Thus section 4 is open for maintenance. After maintenance, the
isolators 3 and 4 are to be closed first and then circuit breaker is
closed.</div>
<div style="text-align: justify;">
In some cases, isolators are used as circuit breaking devices.
But it is limited by particular conditions such are power rating of
given circuit. The isolators are of two types viz single pole and three
pole isolators.</div>
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